Exponential vs Logarithmic Functions Explorer

The grid runs from −5 to 5 on both axes. The teal curve is $f(x) = b^x$, the coral curve is its inverse $f^{-1}(x) = \log_b x$, and the dashed gray diagonal is the line $y = x$. The legend reminds you of the asymptotes: $y = 0$ for the exponential and $x = 0$ for the logarithm.

The readout line shows the pair, for example exp: (1.5, 2.828) ⇄ log: (2.828, 1.5), rounded to three decimals. Notice that the dotted segment always crosses $y = x$ at a right angle: the two points are reflections of each other.

Try b = 1. The exponential becomes the horizontal line $y = 1$, the log curve disappears, and a note says the base must be greater than 0 and not equal to 1. (The readout line still prints a log pair at b = 1; ignore it, because $\log_1 x$ does not exist.) For large bases and large x the highlighted point can sit beyond the edge of the grid, since $5^3 = 125$.

For a base $b > 0$, $b \ne 1$, the logarithm is defined by $$y = \log_b x \quad\Longleftrightarrow\quad b^y = x.$$ So a logarithm answers the question "what exponent on b gives x?" Composing the two functions undoes each one: $\log_b(b^x) = x$ for every real x, and $b^{\log_b x} = x$ for every $x > 0$.

Rates differ, too. For $b > 1$, the exponential's output grows by a constant factor b over equal-length input intervals, while the logarithm's output grows by a constant amount each time the input is multiplied by b. That contrast is how AP Precalculus distinguishes the two models from data.

Step 1: write the inverse. $f^{-1}(x) = \log_2 x$, so we need the exponent y with $2^y = 2.828$.

Step 2: recognize the power. $2.828 \approx 2\sqrt{2} = 2^{1} \cdot 2^{1/2} = 2^{1.5}$. So $\log_2 2.828 \approx 1.5$. (A calculator gives $\ln 2.828 / \ln 2 \approx 1.4998$, the difference coming from rounding 2.828.)

Step 3: match the points. $(1.5,\ 2.828)$ lies on $y = 2^x$ and $(2.828,\ 1.5)$ lies on $y = \log_2 x$. They are reflections across $y = x$.

Check it in the simulator: keep b = 2 and set point x to 1.5. The readout shows exp: (1.5, 2.828) ⇄ log: (2.828, 1.5).

Follow-up with a base below 1. Set b = 0.5 and x = −2. Since $0.5^{-2} = 2^2 = 4$, the readout shows (−2, 4) on the exponential and (4, −2) on the logarithm, so $\log_{0.5} 4 = -2$. Both curves now fall from left to right: the exponential decays toward $y = 0$ and the logarithm drops toward $-\infty$ as x grows.

Topic 2.10 is the bridge into the logarithm half of Unit 2. Expect questions that ask you to read an inverse value from a table of an exponential function, to state the domain or range of a logarithmic function, to convert between $b^y = x$ and $\log_b x = y$, or to choose between an exponential and a logarithmic model from how the outputs change.