Solving Exponential and Logarithmic Equations: Practice

Take the common logarithm of both sides and use the power rule.

$(x + 1)\log 5 = \log 60 \;\Longrightarrow\; x + 1 = \frac{\log 60}{\log 5} \approx \frac{1.77815}{0.69897} \approx 2.54396$

Now subtract $1$, which is the step most often forgotten. To three decimal places, $x \approx 1.544$.

Isolate the power first by dividing both sides by the coefficient $3$.

$4^x = 16 = 4^2 \;\Longrightarrow\; x = 2$

Dividing before matching bases is what keeps the coefficient from tangling with the exponent.

The two arguments force $x > 0$ and $x - 6 > 0$, so the domain is $x > 6$. Combine with the product rule and convert to exponential form.

$\log_2\big(x(x-6)\big) = 4 \;\Longrightarrow\; x(x - 6) = 2^4 = 16$

That gives $x^2 - 6x - 16 = 0$, so $(x - 8)(x + 2) = 0$ and the candidates are $x = 8$ and $x = -2$.

Only $x = 8$ lies in the domain, and it checks out, since $\log_2 8 + \log_2 2 = 3 + 1 = 4$. The candidate $x = -2$ makes both arguments negative, which the combined equation tolerates and the original one does not, so it is extraneous.

The arguments require $x + 3 > 0$ and $x - 1 > 0$, so the domain is $x > 1$. Combine and convert.

$(x + 3)(x - 1) = 5^1 = 5 \;\Longrightarrow\; x^2 + 2x - 8 = 0 \;\Longrightarrow\; (x + 4)(x - 2) = 0$

The candidates are $x = 2$ and $x = -4$. Only $x = 2$ is in the domain, and it checks out, since $\log_5 5 + \log_5 1 = 1 + 0 = 1$. At $x = -4$ both arguments are negative, so that root was manufactured by the combining step.

Convert to exponential form. The statement says that the exponent $2$ on the base $x$ produces $49$.

$x^2 = 49 \;\Longrightarrow\; x = 7 \ \text{ or } \ x = -7$

A logarithm base must be positive and different from $1$, so $x = -7$ is not allowed and the only solution is $x = 7$.

Convert to exponential form and solve the linear equation.

$3x - 2 = 2^3 = 8 \;\Longrightarrow\; 3x = 10 \;\Longrightarrow\; x = \frac{10}{3}$

The argument is $8 > 0$, which is guaranteed here, because $2^3$ is positive.

Convert to exponential form. A one-half power is a square root.

$x = 4^{1/2} = \sqrt{4} = 2$

Checking, $\log_4 2 = \frac12$, because $4^{1/2} = 2$.

The arguments require $x > 0$ (which also makes $x + 6 > 0$). Use the quotient rule, then convert.

$\log_3\frac{x + 6}{x} = 2 \;\Longrightarrow\; \frac{x + 6}{x} = 3^2 = 9 \;\Longrightarrow\; x + 6 = 9x \;\Longrightarrow\; x = \frac{3}{4}$

That value is positive, so it passes the domain test. Checking, $\frac{x+6}{x} = \frac{27/4}{3/4} = 9$, and $\log_3 9 = 2$.

An exponential with a positive base is positive for every real input, so $3^x$ can never be negative.

$3^x > 0 \ \text{ for every real } x, \ \text{ while } \ -9 < 0$

There is nothing to solve, and $\log_3(-9)$ is not a number, so writing it down does not name an answer. The equation has no solution. (The value $x = -2$ gives $3^{-2} = \frac19$, which is positive, not $-9$.)

The base is the same on both sides and $\log_2$ is one-to-one, so set the arguments equal.

$x - 1 = 6 - 2x \;\Longrightarrow\; 3x = 7 \;\Longrightarrow\; x = \frac{7}{3}$

Now test the candidate, because equating arguments is a step that can add roots. At $x = \frac73$ the arguments are $x - 1 = \frac43 > 0$ and $6 - 2x = \frac43 > 0$, both positive, so this one is genuine.

Write the base $\frac12$ as a power of $2$, since $\frac12 = 2^{-1}$.

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

Checking, $\left(\frac12\right)^{-3} = 2^3 = 8$. A base smaller than $1$ needs a negative exponent to grow.

Convert to exponential form. The argument here is $x^2$, not $x$.

$x^2 = 10^2 = 100 \;\Longrightarrow\; x = 10 \ \text{ or } \ x = -10$

Both values are genuine. The domain test asks whether the argument is positive, not whether $x$ is, and $(-10)^2 = 100 > 0$ passes.

Rewriting $\log(x^2)$ as $2\log x$ is what tempts people to lose the negative root, but that rewrite is only valid when $x > 0$.