Solving Exponential and Logarithmic Equations: Additional Practice (Set 1)

The bases match and $\log_7$ is one-to-one, so equate the arguments.

$x^2 + 3x = 10 \;\Longrightarrow\; x^2 + 3x - 10 = 0 \;\Longrightarrow\; (x + 5)(x - 2) = 0$

Now test both candidates, since equating arguments can add roots. At $x = 2$ the left argument is $4 + 6 = 10 > 0$. At $x = -5$ the left argument is $25 - 15 = 10 > 0$ as well.

Both arguments are positive, so both roots are genuine. A negative $x$ is not automatically extraneous: what matters is the sign of the argument, not the sign of $x$.

Isolate the power by dividing both sides by $3$, then match bases.

$2^{x+1} = 32 = 2^5 \;\Longrightarrow\; x + 1 = 5 \;\Longrightarrow\; x = 4$

The exponent is $x + 1$, not $x$, so the final subtraction matters. Checking, $3 \cdot 2^5 = 3 \cdot 32 = 96$.

Convert to exponential form and read the rational exponent as a root followed by a power.

$x = 8^{2/3} = \left(\sqrt[3]{8}\right)^2 = 2^2 = 4$

Checking, $\log_8 4 = \frac23$, because $8^{2/3} = 4$.

The bases agree, so equate the arguments.

$2x - 1 = x + 4 \;\Longrightarrow\; x = 5$

Now test the candidate in the original equation, because equating arguments is a step that can add roots. At $x = 5$ the arguments are $2(5) - 1 = 9$ and $5 + 4 = 9$, both positive, so the solution is genuine and both sides equal $\log_3 9 = 2$.

Because $25^x = \left(5^2\right)^x = \left(5^x\right)^2$, the substitution $u = 5^x$ turns this into a quadratic.

$u^2 - 6u + 5 = 0 \;\Longrightarrow\; (u - 1)(u - 5) = 0 \;\Longrightarrow\; u = 1 \ \text{ or } \ u = 5$

Undo the substitution. From $5^x = 1 = 5^0$ we get $x = 0$, and from $5^x = 5 = 5^1$ we get $x = 1$.

The solutions are $x = 0$ and $x = 1$. The numbers $1$ and $5$ are values of $u$, not of $x$.

The arguments require $x > 0$ and $x + 6 > 0$, so the domain is $x > 0$. Combine and convert.

$x(x + 6) = 4^2 = 16 \;\Longrightarrow\; x^2 + 6x - 16 = 0 \;\Longrightarrow\; (x + 8)(x - 2) = 0$

The candidates are $x = 2$ and $x = -8$, and only $x = 2$ is positive. Checking, $\log_4 2 + \log_4 8 = \frac12 + \frac32 = 2$.

At $x = -8$ the arguments are $-8$ and $-2$, both negative, so the original equation is undefined there and the root is extraneous.

The arguments require $x - 4 > 0$ and $x - 6 > 0$, so the domain is $x > 6$. Move the second logarithm to the left and combine.

$\log_2\big((x - 4)(x - 6)\big) = 3 \;\Longrightarrow\; (x - 4)(x - 6) = 2^3 = 8$

Expanding gives $x^2 - 10x + 16 = 0$, so $(x - 2)(x - 8) = 0$ and the candidates are $x = 2$ and $x = 8$.

Only $x = 8$ satisfies $x > 6$. Checking, $\log_2 4 = 2$ and $3 - \log_2 2 = 3 - 1 = 2$. At $x = 2$ both arguments are negative ($-2$ and $-4$) while their product is $8$, so that candidate satisfies the combined equation and not the original one.

The bases are already the same, and $7^t$ never repeats a value, so the exponents must be equal.

$x + 2 = 3x - 4 \;\Longrightarrow\; 6 = 2x \;\Longrightarrow\; x = 3$

Checking, both exponents become $5$, so both sides are $7^5$.

The arguments require $x > 0$ and $x - 2 > 0$, so the domain is $x > 2$. Use the quotient rule and convert.

$\frac{x}{x - 2} = 5^1 = 5 \;\Longrightarrow\; x = 5x - 10 \;\Longrightarrow\; 4x = 10 \;\Longrightarrow\; x = \frac{5}{2}$

Since $\frac52 > 2$, the candidate is in the domain. Checking, $\frac{5/2}{1/2} = 5$ and $\log_5 5 = 1$.

A single logarithm equal to a constant converts directly to exponential form.

$x^2 - 9 = 2^4 = 16 \;\Longrightarrow\; x^2 = 25 \;\Longrightarrow\; x = 5 \ \text{ or } \ x = -5$

Both roots are genuine. At either value the argument is $25 - 9 = 16$, which is positive, so both satisfy the domain condition. That was guaranteed in advance, because a solution forces the argument to equal $2^4$, a positive number.

Discarding $x = -5$ for being negative would throw away half of the answer. The argument is what must be positive, not $x$.

The bases differ, so take the common logarithm of both sides and use the power rule.

$x \log 3 = (x - 1)\log 5 \;\Longrightarrow\; x \log 3 - x \log 5 = -\log 5 \;\Longrightarrow\; x(\log 5 - \log 3) = \log 5$

By the quotient rule the coefficient is $\log \frac53$, so

$x = \frac{\log 5}{\log \frac53} \approx \frac{0.69897}{0.22185} \approx 3.15066$

To three decimal places, $x \approx 3.151$.

Convert to exponential form. The exponent $4$ on the base $x$ produces $81$.

$x^4 = 81 \;\Longrightarrow\; x = 3 \ \text{ or } \ x = -3$

A logarithm base must be positive and different from $1$, so $x = -3$ is not allowed, even though $(-3)^4 = 81$ is true. The only solution is $x = 3$.