Solving Exponential and Logarithmic Equations: Additional Practice (Set 2)

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

$x^2 - 4x = 3^1 = 3 \;\Longrightarrow\; x^2 - 4x - 3 = 0 \;\Longrightarrow\; x = \frac{4 \pm \sqrt{16 + 12}}{2} = 2 \pm \sqrt{7}$

Both candidates are genuine, and no checking is even required: at any solution the argument equals $3$, which is positive, so the domain condition is satisfied automatically. Converting $\log_b u = k$ into $u = b^k$ can never manufacture a root.

Do not reject $2 - \sqrt{7} \approx -0.646$ for being negative. The test is whether the argument is positive, and it is.

Both $16$ and $8$ are powers of $2$, so rewrite each side on base $2$.

$2^{4x} = 2^{3(x+1)} \;\Longrightarrow\; 4x = 3x + 3 \;\Longrightarrow\; x = 3$

Checking, $16^3 = 2^{12}$ and $8^4 = 2^{12}$.

Take the common logarithm of both sides and bring the exponents down.

$x \log 6 = (x + 2)\log 2 \;\Longrightarrow\; x(\log 6 - \log 2) = 2\log 2$

By the quotient rule, $\log 6 - \log 2 = \log 3$, so

$x = \frac{2 \log 2}{\log 3} \approx \frac{0.60206}{0.47712} \approx 1.26186$

To three decimal places, $x \approx 1.262$. (Exactly, $x = \log_3 4$.)

The arguments require $x > 0$ and $x - 5 > 0$, so the domain is $x > 5$. Combine and convert to exponential form.

$x(x - 5) = 6^1 = 6 \;\Longrightarrow\; x^2 - 5x - 6 = 0 \;\Longrightarrow\; (x - 6)(x + 1) = 0$

The candidates are $x = 6$ and $x = -1$, and only $x = 6$ is in the domain. Checking, $\log_6 6 + \log_6 1 = 1 + 0 = 1$.

At $x = -1$ both arguments are negative while their product is positive, which is exactly the gap the combining step opened up, so that root is extraneous.

The arguments require $x + 2 > 0$ and $x - 5 > 0$, so the domain is $x > 5$. Use the quotient rule and convert.

$\frac{x + 2}{x - 5} = 2^3 = 8 \;\Longrightarrow\; x + 2 = 8x - 40 \;\Longrightarrow\; 42 = 7x \;\Longrightarrow\; x = 6$

Since $6 > 5$, the candidate passes. Checking, $\log_2 8 - \log_2 1 = 3 - 0 = 3$.

Convert to exponential form, remembering that the base is $10$.

$3x - 2 = 10^2 = 100 \;\Longrightarrow\; 3x = 102 \;\Longrightarrow\; x = 34$

The argument is $100 > 0$, as it must be, since $10^2$ is positive. Forgetting to add the $2$ back before dividing is what produces $\frac{100}{3}$.

The arguments require $x > 0$, $x + 2 > 0$, and $x + 6 > 0$, so the domain is $x > 0$. Combine the left side and then equate arguments.

$\log_2\big(x(x + 2)\big) = \log_2(x + 6) \;\Longrightarrow\; x^2 + 2x = x + 6 \;\Longrightarrow\; x^2 + x - 6 = 0$

Factoring gives $(x + 3)(x - 2) = 0$, so the candidates are $x = 2$ and $x = -3$. Only $x = 2$ is in the domain.

Checking $x = 2$: the left side is $\log_2 2 + \log_2 4 = 1 + 2 = 3$ and the right side is $\log_2 8 = 3$. The candidate $x = -3$ is extraneous, so there is exactly one solution.

Put both sides on base $5$, using $25 = 5^2$.

$5^{2x-1} = 5^{2(x-2)} = 5^{2x-4} \;\Longrightarrow\; 2x - 1 = 2x - 4 \;\Longrightarrow\; -1 = -4$

The $x$ terms cancel and a false statement is left, so no value of $x$ works. Looked at another way, the left side is always $5^3 = 125$ times the right side, so the two can never be equal.

The equation has no solution.

Write $81$ as a power of $3$, then equate the exponents.

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

Both roots are genuine. An exponential equation carries no domain restriction, since $3^t$ is defined for every real $t$, so equating exponents cannot manufacture a solution.

Checking $x = -1$: the exponent is $(-1)^2 - 3(-1) = 1 + 3 = 4$, and $3^4 = 81$.

Write both sides on base $2$, using $8 = 2^3$ and $\frac12 = 2^{-1}$.

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

Checking, $8^{2/3 - 1} = 8^{-1/3} = \frac{1}{\sqrt[3]{8}} = \frac12$.

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

$(x - 2)\log 5 = x \log 4 \;\Longrightarrow\; x\log 5 - x \log 4 = 2 \log 5 \;\Longrightarrow\; x(\log 5 - \log 4) = 2\log 5$

The quotient rule turns $\log 5 - \log 4$ into $\log \frac54$, so

$x = \frac{2 \log 5}{\log \frac54} \approx \frac{1.39794}{0.09691} \approx 14.42514$

To three decimal places, $x \approx 14.425$. The answer is large because $\frac54$ is close to $1$, so its logarithm is small and the quotient is big.

Since $4^x = \left(2^x\right)^2$, substitute $u = 2^x$ and factor the quadratic.

$u^2 - u - 56 = 0 \;\Longrightarrow\; (u - 8)(u + 7) = 0 \;\Longrightarrow\; u = 8 \ \text{ or } \ u = -7$

The value $u = -7$ gives no $x$ at all, because $2^x$ is positive for every real $x$ and never reaches $-7$. From $u = 8 = 2^3$ we get $x = 3$.

Checking, $4^3 - 2^3 = 64 - 8 = 56$. The solution is $x = 3$.