Identify the base (2), the exponent (x), and the result (32). Rewrite as $log_2(32) = x$.

Ask: 'To what power must I raise 5 to get 25?' Since $5^2 = 25$, $log_5(25) = 2$.

Convert to exponential form: $3^4 = x$. Then, $x = 81$.

Convert the logarithmic equation to exponential form: $2^5 = x$. Simplify to find $x = 32$.

Recognize that the base is 10. Determine the power of 10 that equals 10000: $10^4 = 10000$. Therefore, $log(10000) = 4$.

Use the property that $ln(e^x) = x$. Therefore, $ln(e^7) = 7$.

Convert to exponential form: $b^y = x$.

Divide both sides by 2: $log(x) = 3$. Convert to exponential form: $10^3 = x$. Therefore, $x = 1000$.

Convert to exponential form: $2^3 = x + 1$. Simplify: $8 = x + 1$. Subtract 1 from both sides: $x = 7$.

Convert to exponential form: $e^0 = x$. Since any number raised to the power of 0 is 1, $x = 1$.

The exponent to which a base must be raised to produce a given number. If $b^a = c$, then $log_b(c) = a$.

The base of a common logarithm is 10. It is written as $log(c)$.

The base of a natural logarithm is $e$ (Euler's number, ≈ 2.71828). It is written as $ln(c)$.

The number you're taking the logarithm of.

The base is the number that is raised to a power to obtain the argument. It must be positive and not equal to 1.

A scale in which units represent a multiplicative change of the base, where each unit is a power of the base.

The exponent to which you raise the base to get the argument.

Euler's Number is the base of the natural logarithm, approximately equal to 2.71828.

The inverse function is $x = log_b(c)$

The argument of $log_b(c)$ is $c$.

The domain is $x > 0$, meaning the graph exists only for positive values of $x$.

The range is all real numbers, meaning $y$ can take any real value.

If $b > 1$, the graph is increasing. If $0 < b < 1$, the graph is decreasing.

There is a vertical asymptote at $x = 0$.

The derivative of $y = e^x$ is also $e^x$, meaning the rate of change is always positive and equal to the function's value.

The derivative of $y = ln(x)$ is $1/x$, which is always positive for $x > 0$.

The derivative of $y = log(x)$ is $\frac{1}{x \cdot ln(10)}$, which is always positive for $x > 0$.

The x-intercept is always at $x = 1$, since $log_b(1) = 0$ for any valid base $b$.

The graph of $y = log_b(x)$ is concave down for $b > 1$ and concave up for $0 < b < 1$.

As $x$ approaches infinity, $y = log_b(x)$ also approaches infinity, but at a slower rate than a linear function.