# How to find the equation of a logarithm function from its graph? (2022)

First, let's revise some of the "typical" shapes a logarithm curve can take. If you're not sure about the background, see the chapter Exponential and Logarithmic Functions.

A logarithm graph of the form y = log10(x) has the following shape:

Figure 1: The graph of y = log10(x)

Notice the graph passes through the point (1, 0) (since
log(1) = 0, no matter what base we are using) and the point (10, 1), since log10(10) = 1.

## The graph of log x using different bases

The following graph shows the cases when the base is:

• 2 (the top-most curve), passing through:
• (4, 2) since log24 = 2, that is, 22 = 4 and
• (8, 3) since log28 = 3, that is 23 = 8;
• e ≈ 2.71828 (the magenta curve);
• 3, passing though (9, 2) since 32 = 9
• 4, 5, 6, 7, 8, 9; and finally
• base 10 (dark green).

Figure 2: The graph of y = logb(x) for various values of b.

## Finding the base from the graph

We can find the base of the logarithm as long as we know one point on the graph. Here, we assume the curve hasn't been shifted in any way from the "standard" logarithm curve, which always passes through (1, 0).

Example: A logarithmic graph, y = logb(x), passes through the point (12, 2.5), as shown. What is the base, b?

Example 1: Find the base

Answer: We substitute in our known values y = 2.5 when x = 12.

2.5 = logb(12)

To find the base, we just need to apply the basic logarithm identity:

If y = logb(x), then

by = x

Applying this, we have:

b2.5 = 12

This gives

(Video) How to Write Equation of Logarithmic Function From Graph MHF4U Pre Calculus

So the base of the given logarithm equation is 2.7.

Let's now see some "non-standard" ways the logarithm graph can appear.

## Multiplying the log term

The general form for this curve is:

y = d log10(x)

If we multiply the log term, we elongate (or compress) the graph in the vertical direction. In this next graph, we see from the top-most curve:

d = 5, 4, 3, 2;

y = log10(x) (dark green);

y = 0.5 log10(x)

Figure 3: Graph of y = d log10(x) for various values of d

While this may look similar to Figure 2 above, it is quite a different situation.

## Reflecting the log graph in the y-axis

Here is the graph of y = log10(−x).

The minus sign before the "x" has the effect of reflecting the curve in the y-axis. We get a vertical mirror image of the curve we saw in Figure 1.

Figure 4: Graph of y = log10(−x)

## Reflecting the log graph in the x-axis

Here is the graph of y = log10(x).

The minus sign before the "log(x)" has the effect of reflecting the curve in the x-axis. We get a mirror image horizontally.

Figure 5: Graph of y = log10(x)

## Reflecting the log graph in both the x-axis and y-axis

Here is the graph of y = log10(x).

(Video) Example Finding a log equation given the graph

We get a mirror image across both the vertical and horizontal axes.

Figure 6: Graph of y = log10(x)

NOTE: Compare Figure 6 to the graph we saw in Graphs of Logarithmic and Exponential Functions, where we learned that the exponential curve is the reflection of the logarithmic function in the line y = x. This is not the same situation as Figure 1 compared to Figure 6.

Next, we'll see what happens when we come across logarithm graphs that do not pass neatly though (1, 0) or (−1, 0) as in the cases we've just seen.

## Shifting the logarithm function up or down

We introduce a new formula,

y = c + log(x)

The c-value (a constant) will move the graph up if c is positive and down if c is negative.

For example, here is the graph of y = 2 + log10(x). Notice it passes through (1, 2). It is the curve in Figure 1 shifted up by 2 units.

Example 2: Graph of y = 2 + log10(x).

Similarly, this is the graph of y = −2 + log10(x). It is the result of shifting the curve in Figure 1 down by 2 units, and it passes through (1, −2).

Example 3: Graph of y = −2 + log10(x).

## Shifting the logarithm function left or right

We introduce another new formula, where we've replaced x with (x + a):

y = log(x + a)

The "a" will have the effect of shifting the logarithm curve right if a is negative, and to the left if a is positive.

For example, the graph of y = log10(x − 4) looks like:

Example 4: Graph of y = log10(x − 4)

The curve in Figure 1 has been shifted to the right by 4 units.

Also, the graph of y = log10(x + 5) gives us the curve in Figure 1 moved left by 5 units:

(Video) Write Equation of Logarithmic Function from Graph

Example 5: Graph of y = log10(x + 5)

## Summary example

Let's pull what we've learned so far into an example.

Question: Find the equation of the logarithm function (base 10) for the following curve.

Example 6: Find the logarithmic function

Answer: We observe the shape of this curve to be closest to Figure 4, which was y= log10(−x).

We'll assume the general equation is:

y= c + log10(−x + a).

We also observe the (almost) vertical portion of the graph is at x = 2.5, so we replace −x with (x 2.5) and conclude a = 2.5. (Try a few values for x to see why a ends up being positive.)

Here's an interim graph, where I've moved the curve 2.5 units left so I can easily see how high or low the graph is:

Example 6a: Interim graph, moved 2.5 units left

We know the graph needs to pass through (−1, 0), and we observe we are 1.5 units too high. So we conclude c = 1.5.

So the required equation is

y= 1.5 + log10(−x + 2.5).

## Determining the equation of a logarithmic function from data

There are many computer packages (SPSS, Excel, etc) that can determine a "best fit" curve for a given set of data. But for the sake of explaining how to determine an unknown logarithmic function from its data plot, let's take a look at this example (one which is close to my heart).

The experiment: A group of people are asked to learn a list of random words. They are tested immediately, then again after some time has elapsed, and then repeatedly over longer time spans. As expected, the number of words they remember correctly diminishes over time.

The average data is plotted over time as follows, where the horizontal axis is time, and the vertical axis is the proportion of the words they got right. (This is the famous Ebbinghaus Forgetting Curve.)

Example 7: Ebbinghaus Forgetting data

(Video) Finding The Equation of a Logarithm Function From its Graph (Without a Calculator)

The question: You can see a "best fit" curve has been drawn through the data points. What is the equation of that curve?

The answer: We observe it is most like Figure 5 , which had the formula y= log10(x).

We assume our logarithmic function will have the form:

y= c − d log10(x).

We now aim to find the values of c and d.

We know from the section on Graphs on Logarithmic and Semi-Logarithmic Axes that we can turn a logarithmic (or exponential) curve into a linear curve by taking the logarithm of one of the variables.

To do so, we take the original data set (the columns "time" and "proportion") and find the logarithm of the independent variable, time. (Normally we would use natural log for such analysis, but I'm sticking with base 10 to be consistent with the earlier examples.)

time
(t)
log10(t)proportion
100.84
50.698970.71
151.176090.61
301.477120.56
601.778150.54
1202.079180.47
2402.380210.45
4802.681240.38
7202.857330.36
14403.158360.26
28803.459390.2
57603.760420.16
100804.003460.08

Now we can easily deduce the linear equation for this curve (I'm taking the first data point and the second last, since the last one is clearly not that close to the smooth curve.)

I use the point-slope formula for a line:

y= 0.79872 0.16985 x

Next, we simply replace the "x" with log10(x) and achieve the required equation:

y= 0.79872 0.16985 log10(x)

NOTE: The Ebbinghaus Forgetting curve is usually modeled using an exponential curve, but this data quite nicely fits a logarithmic curve for the given values. It's not a good model if we extrapolate it into the future, since this logarithmic curve doesn't flatten out (like the appropriate exponential one would, and it eventually goes off into negative territory.)

[Data credit: Penn State]

## Conclusion

Hopefully, whatever logarithmic graph you are trying to find the equation for will be covered by one or more of the cases above.

(Video) Given the equation of a logarithmic function, identify its graph

## FAQs

### How will you determine the equation of a logarithmic function given its graph? ›

Forming an Equation from a Log Graph - YouTube

### How do you find the logarithmic equation? ›

The logarithmic function for x = 2y is written as y = log2 x or f(x) = log2 x. The number 2 is still called the base. In general, y = logb x is read, “y equals log to the base b of x,” or more simply, “y equals log base b of x.” As with exponential functions, b > 0 and b ≠ 1.
...
x = 3yy
−1
10
31
92
1 more row

### How do you find the equation of a logarithm with two points? ›

Finding Equations from Logarithmic and Exponential Graphs - YouTube

### How do you find the logarithmic equation from a table? ›

How to use a table to graph logarithmic function - YouTube

### How do you match a logarithmic function with a graph? ›

Ex 1: Match Graphs with Exponential and Logarithmic Functions - YouTube

### How do you graph log functions step by step? ›

Graphing Logarithmic Functions
1. Step 1: Find some points on the exponential f(x). The more points we plot the better the graph will look.
2. Step 2: Switch the x and y values to obtain points on the inverse.
3. Step 3: Determine the asymptote.
4. Graph the following logarithmic functions. State the domain and range.

If you have a single logarithm on each side of the equation having the same base, you can set the arguments equal to each other and then solve.. If you have a single logarithm on one side of the equation, you can express it as an exponential equation and solve it.. Since we want to transform the left side into a single logarithmic equation, we should use the Product Rule in reverse to condense it.. However, it is NOT ALLOWED to have a logarithm of a negative number or a logarithm of zero, 0 , when substituted or evaluated into the original logarithm equation.. We want to have a single log expression on each side of the equation.. What we have here are differences of logarithmic expressions on both sides of the equation.. I think we are ready to set each argument equal to each other since we can reduce the problem to have a single log expression on each side of the equation.. On the left side, we see a difference of logs which means we apply the Quotient Rule while the right side requires the Product Rule because they’re the sum of logs.. When you check x=0 back into the original logarithmic equation, you’ll end up having an expression that involves getting the logarithm of zero, which is undefined, meaning – not good!. Collect all the logarithmic expressions on one side of the equation (keep it on the left) and move the constant to the right side.. When you check x=1 back to the original equation, you should agree that \large{\color{blue}x=1} is the solution to the log equation.. At this point, I used different colors to illustrate that I’m ready to express the log equation into its exponential equation form.

The graph of a quadratic function is a parabola.. We know that a quadratic equation will be in the form:. This is a quadratic function which passes through the x -axis at the required points.. Let's substitute x = 0 into the equation I just got to check if it's correct.. So how do we find the correct quadratic function for our original question (the one in blue)?. To find the unique quadratic function for our blue parabola, we need to use 3 points on the curve.. So the correct quadratic function for the blue graph is. For this parabola, the vertex is at ( h, k ).. In this example, the blue curve passes through (0, 1) on the y -axis, so we can simply substitute x = 0, y = 1 into y = a ( x − 1) 2 as follows:. We just substitute as before into the vertex form of our quadratic function.. GeoGebra will give us the equation of a parabola, but you need to know the focus and directrix first.

The function f (x) = log b x is read as “log base b of x.” Logarithms are useful in mathematics because they enable us to perform calculations with very large numbers.. Properties of logarithmic functions are simply the rules for simplifying logarithms when the inputs are in the form of division, multiplication, or exponents of logarithmic values.. The product rule of logarithm states the logarithm of the product of two numbers having a common base is equal to the sum of individual logarithms.. The quotient rule of logarithms states that the logarithm of the two numbers’ ratio with the same bases is equal to the difference of each logarithm.. The power rule of logarithm states that the logarithm of a number with a rational exponent is equal to the product of the exponent and its logarithm.. Exponential functionLogarithmic functionRead as 8 2 = 64log 8 64 = 2log base 8 of 6410 3 = 1000log 1000 = 3log base 10 of 100010 0 = 1log 1 = 0log base 10 of 125 2 = 625log 25 625 = 2log base 25 of 62512 2 = 144log 12 144 = 2log base 12 of 144Let’s use these properties to solve a couple of problems involving logarithmic functions.. ⟹ log 2 + log 5 = log (2 * 5) = Log(10).

f is a function given by. f (x) = log 2 (x + 2) Find the domain and range of f. Find the vertical asymptote of the graph of f. Find the x and y intercepts of the graph of f if there are any.. Matched Problem to Example 1 f is a function given by. f (x) = log 2 (x + 3) Find the domain of f and range of f. Find the vertical asymptote of the graph of f. Find the x and y intercepts of the graph of f if there are any.. f is a function given by. f (x) = - 3 ln(x - 4) Find the domain of f and range of f. Find the vertical asymptote of the graph of f. Find the x and y intercepts of the graph of f if there are any.. Matched Problem to Example 2 f is a function given by Find the domain of f and range of f. Find the vertical asymptote of the graph of f. Find the x and y intercepts of the graph of f if there are any.. f is a function given by. f (x) = 2ln(| x |) Find the domain of f and range of f. Find the vertical asymptote of the graph of f. Find the x and y intercepts of the graph of f if there are any.

Graph logarithmic functions.. shifts the parent function left units if shifts the parent function right units if has the vertical asymptote has domain has range. shifts the parent function up units if shifts the parent function down units if has the vertical asymptote has domain has range. The function. The function. Given a logarithmic function with the parent function graph a translation.. Graphing a Reflection of a Logarithmic Function. Press [GRAPH] to observe the graphs of the curves and use [WINDOW] to find an appropriate view of the graphs, including their point(s) of intersection.. where the parent function,is. Finding the Vertical Asymptote of a Logarithm Graph. For example, look at the graph in .. To find the domain of a logarithmic function, set up an inequality showing the argument greater than zero, and solve for See (Figure) and (Figure) The graph of the parent function has an x- intercept at domain range vertical asymptote and if the function is increasing.

The graph below shows this point.. $f\left(x\right)={2}^{x}$ has a y -intercept at $\left(0,1\right)$ and $g\left(x\right)={\mathrm{log}}_{2}\left(x\right)$ has an x -intercept at $\left(1,0\right)$.. one-to-one function vertical asymptote: x = 0 domain: $\left(0,\infty \right)$ range: $\left(-\infty ,\infty \right)$ x- intercept: $\left(1,0\right)$ and key point $\left(b,1\right)$ y -intercept: none increasing if $b>1$ decreasing if 0 < b < 1. Graph $f\left(x\right)={\mathrm{log}}_{5}\left(x\right)$.. The x -intercept is $\left(1,0\right)$.. Graph $f\left(x\right)={\mathrm{log}}_{\frac{1}{5}}\left(x\right)$.. shifts the parent function $y={\mathrm{log}}_{b}\left(x\right)$ left c units if c > 0. shifts the parent function $y={\mathrm{log}}_{b}\left(x\right)$ right c units if c < 0. has the vertical asymptote x = – c . has domain $\left(-c,\infty \right)$.. The x -intercept will be $\left(-1,0\right)$.. The function $f\left(x\right)={\mathrm{-log}}_{b}\left(x\right)$. The function $f\left(x\right)={\mathrm{log}}_{b}\left(-x\right)$. The x -intercept is $\left(-1,0\right)$.. Graph $f\left(x\right)=-\mathrm{log}\left(-x\right)$.. Press [GRAPH] to observe the graphs of the curves and use [WINDOW] to find an appropriate view of the graphs, including their point(s) of intersection.. General Form for the Transformation of the Parent Logarithmic Function $\text{ }f\left(x\right)={\mathrm{log}}_{b}\left(x\right)$$f\left(x\right)=a{\mathrm{log}}_{b}\left(x+c\right)+d$ To find the domain of a logarithmic function, set up an inequality showing the argument greater than zero and solve for x .. The graph of the parent function $f\left(x\right)={\mathrm{log}}_{b}\left(x\right)$ has an x- intercept at $\left(1,0\right)$, domain $\left(0,\infty \right)$, range $\left(-\infty ,\infty \right)$, vertical asymptote x = 0, and if b > 1, the function is increasing.

${4^2} = 16$. we must have the following value of the logarithm.. Therefore, the value of this logarithm is,. Notice that with this one we are really just acknowledging a change of notation from fractional exponent into radical form.. So, when evaluating logarithms all that we’re really asking is what exponent did we put onto the base to get the number in the logarithm.. Properties of Logarithms $${\log _b}1 = 0$$.. We should also give the generalized version of Properties 3 and 4 in terms of both the natural and common logarithm as we’ll be seeing those in the next couple of sections on occasion.. We’ll start with the common logarithm form of the change of base.

Given a logarithmic function with the form graph the function.. shifts the parent function left units if shifts the parent function right units if has the vertical asymptote has domain has range. Graphing a Vertical Shift of y = log b ( x ) When a constantis added to the parent functionthe result is a vertical shiftunits in the direction of the sign onTo visualize vertical shifts, we can observe the general graph of the parent functionalongside the shift up,and the shift down,See (Figure) .. shifts the parent function up units if shifts the parent function down units if has the vertical asymptote has domain has range. The function. The function. Given a logarithmic function with the parent function graph a translation.. Press [GRAPH] to observe the graphs of the curves and use [WINDOW] to find an appropriate view of the graphs, including their point(s) of intersection.. [reveal-answer q=”fs-id1165135193432″]Show Solution[/reveal-answer]. [hidden-answer a=”fs-id1165135193432″]Press [Y=] and enternext to Y 1 =.. Yes, if we know the function is a general logarithmic function.. For example, look at the graph in .. To find the domain of a logarithmic function, set up an inequality showing the argument greater than zero, and solve for See (Figure) and (Figure) The graph of the parent function has an x- intercept at domain range vertical asymptote and if the function is increasing.

In this section we will now take a look at solving logarithmic equations, or equations with logarithms in them.. In particular we will look at equations in which every term is a logarithm and we also look at equations in which all but one term in the equation is a logarithm and the term without the logarithm will be a constant.. \begin{align*}x & = 6x - 1\\ 1 & = 5x\hspace{0.25in} \Rightarrow \hspace{0.25in}x = \frac{1}{5}\end{align*}. Now, we do need to worry if this solution will produce any negative numbers or zeroes in the logarithms so the next step is to plug this into the original equation and see if it does.. \begin{align*}\log 6 + \log \left( {6 - 1} \right) & = \log \left( {3\left( 6 \right) + 12} \right)\\ \log 6 + \log 5 & = \log 30\end{align*}. No logarithms of negative numbers and no logarithms of zero so this is a solution.. \begin{align*}\ln 10 - \ln \left( {7 - 2} \right) & = \ln 2\\ \ln 10 - \ln 5 & = \ln 2\end{align*}. \begin{align*}\ln 10 - \ln \left( {7 - 5} \right) & = \ln 5\\ \ln 10 - \ln 2 & = \ln 5\end{align*}. In this case both possible solutions, $$x = 2$$ and $$x = 5$$, end up actually being solutions.. \begin{align*}\log 5 & = 1 - \log \left( {5 - 3} \right)\\ \log 5 & = 1 - \log 2\end{align*}. No negative numbers or zeroes in the logarithms and so this is a solution.. We exclude a potential solution if it produces negative numbers or zeroes in the logarithms upon substituting it into the equation and we include a potential solution if it doesn’t.. \begin{align*}{x^2} - 6x & = 8\left( {1 - x} \right)\\ {x^2} - 6x & = 8 - 8x\\ {x^2} + 2x - 8 & = 0\\ \left( {x + 4} \right)\left( {x - 2} \right) & = 0\hspace{0.25in} \Rightarrow \hspace{0.25in}x = - 4,\,\,x = 2\end{align*}. Now, let’s check both of these solutions in the original equation.. \begin{align*}{\log _2}\left( {{{\left( { - 4} \right)}^2} - 6\left( { - 4} \right)} \right) & = 3 + {\log _2}\left( {1 - \left( { - 4} \right)} \right)\\ {\log _2}\left( {16 + 24} \right) & = 3 + {\log _2}\left( 5 \right)\end{align*}. So, upon substituting this solution in we see that all the numbers in the logarithms are positive and so this IS a solution.. \begin{align*}{\log _2}\left( {{2^2} - 6\left( 2 \right)} \right) & = 3 + {\log _2}\left( {1 - 2} \right)\\ {\log _2}\left( {4 - 12} \right) & = 3 + {\log _2}\left( { - 1} \right)\end{align*}. In this case, despite the fact that the potential solution is positive we get negative numbers in the logarithms and so it can’t possibly be a solution.

A logarithmic equation is an equation which involves at least one unknown variable, where a logarithmic expression appears in at least one side of the equation .. $\ln(x-1) = \ln(2y + 1)$ The first disclaimer is that there are no bullet-proof ways of solving a logarithmic equation, nor a general equation for that matter.. First, try to group all logarithmic expression into one logarithmic expression.. So, in other words, solving a logarithmic equation consists of grouping the logarithmic expressions, eliminating them by applying exponential, and then solve the equation as a regular equation.. Solve the following logarithmic equation:. Indeed, eliminating the logarithm is the easy part of solving log equations.

We can find the formula of an exponential function by using two points on the curve, substituting them into the formula y = ab x , and solving the system of two equations in two unknowns.. Of course, we cannot find the formula of an exponential function from one point, since one point (one equation) does not give us enough information to solve for two parameters (the coefficient a and the base b).. Example 1: How To Find The Formula Of An Exponential Function With Two Points Let’s say that you are given the points (0, 1) and (2, 9), which lie on the graph of an exponential function.. Example 2: How To Find The Formula Of An Exponential Function With Two Points Let’s say that you are given the points (1, 10) and (3, 40), which lie on the graph of an exponential function.. Example 3: How To Find The Formula Of An Exponential Function Given A Table Let’s say that you are given the following table of points, which lie on the graph of an exponential function.A table ofvalues for anexponentialfunction.. Example 4: How To Find The Formula Of An Exponential Function Given A Table Let’s say that you are given the following graph of an exponential function.. Example 5: How To Find The Base Of An Exponential Function (From Two Points) Let’s say we have the points (2, 98) and (3, 686) on an exponential function.. Now that we have b = 7, we can use the general formula for an exponential function and the point (2, 98) to find a:. Now you know how to find the formula of an exponential function from two points.

·Convert logarithmic equations to exponential equations.. ·Convert exponential equations to logarithmic equations.. The base of the exponential equation, 11, is also the base (the small subscript number at the end of “log”) in the logarithmic equation.. The base of the exponential equation, 11, is also the base (the small subscript number at the end of “log”) in the logarithmic equation.. The blue graph is the logarithmic function, and the red graph is the corresponding exponential function.. Convert the logarithmic equation to an exponential equation.. Convert the logarithmic equation to an exponential equation.. Convert the logarithmic equation to an exponential equation.. Converting a logarithmic equation to the equivalent exponential equation is helpful with both graphing and solving equations.

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