Solving Exponential and Logarithmic Equations: Practice

Write the right side as a power of the same base. Since $64 = 2^6$, both sides sit on base $2$.

$2^x = 2^6 \;\Longrightarrow\; x = 6$

Equal powers of the same base force equal exponents, because $2^x$ never takes the same value twice.

Rewrite $125$ as a power of $5$, since $5 \cdot 5 \cdot 5 = 125$.

$5^x = 5^3 \;\Longrightarrow\; x = 3$

The bases now match, so the exponents must match.

First put the right side on base $3$, using $27 = 3^3$.

$3^{x+1} = 3^3 \;\Longrightarrow\; x + 1 = 3 \;\Longrightarrow\; x = 2$

Checking, $3^{2+1} = 3^3 = 27$.

Both sides are already powers of $2$, so the exponents must be equal.

$3x = 12 \;\Longrightarrow\; x = 4$

The base is dropped, not divided out, so $12$ is not the answer.

Rewrite the equation in exponential form. The exponent $5$ on the base $2$ produces $x$.

$x = 2^5 = 32$

Multiplying the base by the logarithm, which would give $10$, is not what a logarithm means.

Convert to exponential form, remembering that $5^1 = 5$.

$2x = 5^1 = 5 \;\Longrightarrow\; x = \frac{5}{2}$

The argument is then $2x = 5 > 0$, so the solution stands.

Raise the base $b$ to each side of the equation and use the cancellation identity $b^{\log_b M} = M$, which holds for every positive $M$.

$b^{\log_b M} = b^{\log_b N} \;\Longrightarrow\; M = N$

This works because $\log_b$ is one-to-one, so equal logarithms can only come from equal arguments.

Write the right side as a power of $4$. A reciprocal is a negative exponent, and $16 = 4^2$.

$\frac{1}{16} = \frac{1}{4^2} = 4^{-2} \;\Longrightarrow\; 4^x = 4^{-2} \;\Longrightarrow\; x = -2$

The value of an exponential is never negative, but its exponent certainly can be.

Count the factors of $10$ in $1000$, which is $10 \cdot 10 \cdot 10$.

$10^x = 10^3 \;\Longrightarrow\; x = 3$

In the language of common logarithms, this says $\log 1000 = 3$.

The logarithm is one-to-one, so two logarithms with the same base are equal only when their arguments are equal.

$\log_3 x = \log_3 7 \;\Longrightarrow\; x = 7$

The argument $7$ is positive, so the answer survives the domain check.

Both sides are powers of $3$, since $9 = 3^2$.

$\left(3^2\right)^x = 3^1 \;\Longrightarrow\; 3^{2x} = 3^1 \;\Longrightarrow\; 2x = 1 \;\Longrightarrow\; x = \frac{1}{2}$

The exponent $\frac12$ is a square root, and indeed $\sqrt{9} = 3$.

Write $49$ as a power of $7$, then equate the exponents.

$7^{2x} = 7^2 \;\Longrightarrow\; 2x = 2 \;\Longrightarrow\; x = 1$

Checking, $7^{2(1)} = 49$.