Why These Graphs Fit Together
An exponential function and its logarithmic inverse describe the same relationship from opposite directions. The exponential function takes an input and produces a power. Its inverse takes that power as an input and returns the original exponent.
For instance, \(3^2=9\) says that input \(2\) produces output \(9\) under the exponential function. The inverse relationship says that input \(9\) produces output \(2\) under the logarithm. On a coordinate plane, the corresponding points are \((2,9)\) and \((9,2)\). Their coordinates have exchanged places.
Exchanging the coordinates of every point on a graph is exactly what reflection across the line \(y=x\) does. This tutorial develops that connection and uses it to match the points and key features of exponential and logarithmic graphs.
The Functions and Their Domains
Let \(b\) be a positive real number with \(b\ne 1\). The exponential function with base \(b\) is \(f(x)=b^x\), with domain all real numbers and range the positive real numbers. Its inverse is the logarithmic function \(f^{-1}(x)=\log_b(x)\), whose domain is the positive real numbers and whose range is all real numbers.
Because these functions are inverses, they undo one another on their respective domains:
The condition \(b>0\) and \(b\ne1\) matters. A positive base ensures real powers are defined across the real domain used here; excluding \(b=1\) ensures the exponential function is not constant and can have an inverse. Both bases between zero and one and bases greater than one are allowed.
Reflection Across \(y=x\)
A point \((a,c)\) reflected across the line \(y=x\) becomes \((c,a)\). This follows because the reflection exchanges the horizontal and vertical coordinates. The line itself consists of points whose coordinates are equal, so points on \(y=x\) stay fixed.
Here is a complete justification. A point \((a,c)\) lies on the graph of \(f\) exactly when \(c=f(a)\). By the definition of an inverse, \(c=f(a)\) exactly when \(a=f^{-1}(c)\). That last equality says that \((c,a)\) lies on the graph of \(f^{-1}\). Therefore each point on the graph of \(f\) corresponds to a point on the inverse graph with its coordinates exchanged. The same reasoning works in reverse, so the correspondence includes every point on both graphs. Exchanging coordinates is reflection across \(y=x\), as claimed.
For the exponential function, the inverse relationship can also be seen directly. If \((x,b^x)\) is on \(y=b^x\), then \(\log_b(b^x)=x\), so \((b^x,x)\) is on \(y=\log_b(x)\). Conversely, if \((u,\log_b(u))\) is on the logarithm graph, then \(b^{\log_b(u)}=u\), so \((\log_b(u),u)\) is on the exponential graph.
Worked Example: Reflecting Points for Base 3
Worked Example: Reflecting Points for Base 3
Find several points on \(y=3^x\), reflect them to \(y=\log_3(x)\), and check the matching logarithm values.
Evaluate the exponential function at \(x=-2,-1,0,1,2\):
The resulting exponential points are \((-2,\frac19)\), \((-1,\frac13)\), \((0,1)\), \((1,3)\), and \((2,9)\). Swap the coordinates of each point to obtain the logarithm points: \((\frac19,-2)\), \((\frac13,-1)\), \((1,0)\), \((3,1)\), and \((9,2)\).
Check the correspondence using the definition of logarithm: \(\log_3(\frac19)=-2\), since \(3^{-2}=\frac19\); \(\log_3(\frac13)=-1\), since \(3^{-1}=\frac13\); \(\log_3(1)=0\), since \(3^0=1\); \(\log_3(3)=1\), since \(3^1=3\); and \(\log_3(9)=2\), since \(3^2=9\).
Each logarithm value agrees with the corresponding reflected point. For example, the exponential point \((2,9)\) reflects to \((9,2)\), and \(\log_3(9)=2\). The check confirms the coordinate swap rather than merely suggesting it.
Worked Example: A Decreasing Pair of Graphs
Worked Example: A Decreasing Pair of Graphs
Use base \(b=\frac13\) to find exponential and logarithmic points, then compare the order of the logarithm values as its inputs increase.
For \(y=(\frac13)^x\), evaluate at \(x=-2,-1,0,1,2\):
The exponential points are \((-2,9)\), \((-1,3)\), \((0,1)\), \((1,\frac13)\), and \((2,\frac19)\). Swapping coordinates gives logarithm points \((9,-2)\), \((3,-1)\), \((1,0)\), \((\frac13,1)\), and \((\frac19,2)\).
To read the logarithm graph from left to right, put the input coordinates in increasing order: \(\frac19,\frac13,1,3,9\). Their logarithm values are, respectively, \(2,1,0,-1,-2\). Thus as \(x\) increases from \(\frac19\) to \(\frac13\), to \(1\), to \(3\), to \(9\), the logarithm's values decrease from \(2\) to \(1\), to \(0\), to \(-1\), to \(-2\).
Check each value from the exponential definition: \((\frac13)^2=\frac19\), \((\frac13)^1=\frac13\), \((\frac13)^0=1\), \((\frac13)^{-1}=3\), and \((\frac13)^{-2}=9\). Therefore the listed logarithm values are correct, and the graph is decreasing as its input increases.
How Graph Features Correspond
Reflection exchanges horizontal and vertical information. A graph's domain, which describes its allowed horizontal coordinates, becomes the inverse graph's range, which describes its vertical coordinates. Likewise, the original range becomes the inverse domain.
For \(y=b^x\), the domain is \(\mathbb R\), and the range is \((0,\infty)\). Therefore, for \(y=\log_b(x)\), the domain is \((0,\infty)\), and the range is \(\mathbb R\). The logarithm is not defined for zero or negative inputs, in agreement with the fact that \(b^x\) is always positive.
The exponential graph crosses the vertical axis at \((0,1)\), because \(b^0=1\). Reflection sends that point to \((1,0)\), the horizontal-axis intercept of the logarithm graph, because \(\log_b(1)=0\). These intercepts are a useful quick check when sketching either graph.
The exponential graph has horizontal asymptote \(y=0\). Under reflection, this horizontal line becomes the vertical line \(x=0\), which is the logarithm graph's vertical asymptote. The logarithm graph approaches this line from the right because its domain only includes positive inputs. These asymptotes are another pair of corresponding features, not unrelated facts.
The direction of the graph is also preserved as a type of behavior. When \(b>1\), the exponential graph increases, and the logarithm graph increases. When \(0<b<1\), both decrease. Reflection swaps the roles of input and output but does not turn an increasing inverse pair into one increasing and one decreasing.
Worked Example: Matching Intercepts and Asymptotes
Worked Example: Matching Intercepts and Asymptotes
Describe the corresponding intercepts and asymptotes for \(y=2^x\) and \(y=\log_2(x)\).
First, \(2^0=1\), so \((0,1)\) lies on \(y=2^x\). Reflecting across \(y=x\) exchanges the coordinates and gives \((1,0)\). Check it on the logarithm graph: \(\log_2(1)=0\), because \(2^0=1\). Thus \((0,1)\) is the exponential y-intercept and \((1,0)\) is the logarithmic x-intercept.
Next, the exponential graph has horizontal asymptote \(y=0\). Swapping the coordinates in this line gives \(x=0\), the vertical asymptote for the logarithm graph. For \(b=2\), as the exponential output approaches \(0\) from above, the corresponding input on the logarithm graph approaches \(0\) from the right. The logarithm graph therefore approaches \(x=0\) without including \(x=0\) in its domain.
The checks are consistent: \(\log_2(1)=0\) verifies the reflected intercept, and the logarithm's domain \(x>0\) verifies that its graph can approach \(x=0\) only from the right.
A Common Mistake: Taking Reciprocals
A frequent error is to confuse exchanging coordinates with taking a reciprocal. Reflection across \(y=x\) sends \((a,c)\) to \((c,a)\); it does not send the point to \((a,\frac1c)\).
The correction is to swap coordinates: \((1,3)\) becomes \((3,1)\). Check the logarithm graph: \(\log_3(3)=1\), so \((3,1)\) is indeed on \(y=\log_3(x)\). The reciprocal value \(\frac13\) is related to another power, \(3^{-1}=\frac13\), but it does not describe the reflection of the point \((1,3)\).
A Reliable Sketching Procedure
When sketching either member of an inverse pair, use the same steps. The point correspondence provides a direct way to transfer a sketch, while the domains, intercepts, and asymptotes help check that the result is consistent.
Check that \(b>0\) and \(b\ne1\), so \(y=b^x\) and \(y=\log_b(x)\) are inverse functions.
For example, use \(b^0=1\) and \(b^1=b\) to obtain exponential points \((0,1)\) and \((1,b)\).
The corresponding logarithm points are \((1,0)\) and \((b,1)\). Each new point should satisfy the logarithm definition.
Exchange domain and range, carry the exponential horizontal asymptote to the logarithm's vertical asymptote, and check the intercepts.
Why the Reflection Matters
The reflection is more than a visual pattern. It lets you use information about one graph to understand its inverse without calculating the inverse's points from scratch. If a point on the exponential graph is known, its coordinates immediately give a point on the logarithm graph. If a graph feature depends on horizontal coordinates, reflection indicates how it appears in the vertical direction, and vice versa.
The method also provides checks against errors. An exponential output must be positive, so the logarithm's domain must be positive. The exponential y-intercept \((0,1)\) must correspond to the logarithm x-intercept \((1,0)\). If a proposed sketch violates one of these correspondences, it cannot be the correct inverse graph.
Keep the central rule in view: inverse-function graphs reflect across \(y=x\) because each input and output trade roles. For exponential and logarithmic functions, this rule explains their paired points, their domain-range exchange, and the correspondence between their intercepts and asymptotes.
Check Your Understanding
Use the inverse relationship and coordinate swapping to answer these questions.
- If \((2,16)\) lies on \(y=4^x\), what point lies on \(y=\log_4(x)\), and what logarithm equation checks it?
- For \(y=(\frac12)^x\), calculate the outputs at \(x=-1,0,1\), then write the reflected points on \(y=\log_{\frac12}(x)\).
- What are the domain and range of \(y=\log_b(x)\) when \(b>0\) and \(b\ne1\), and how do they relate to those of \(y=b^x\)?
- Why does the exponential y-intercept correspond to the logarithm x-intercept?
- A student replaces a point \((a,c)\) on an exponential graph by \((a,\frac1c)\). Explain the error and state the correct reflected point.
