Solving Exponential and Logarithmic Equations: Practice

The bases $2$ and $3$ share no common base, so take the common logarithm of both sides and use the power rule.

$(x + 1)\log 2 = (x - 1)\log 3$

Expand and gather the $x$ terms on one side: $x\log 2 + \log 2 = x\log 3 - \log 3$, so $\log 2 + \log 3 = x(\log 3 - \log 2)$. The product and quotient rules tidy both sides.

$x = \frac{\log 6}{\log \frac32} \approx \frac{0.77815}{0.17609} \approx 4.41903$

To three decimal places, $x \approx 4.419$.

Since $9^x = \left(3^2\right)^x = \left(3^x\right)^2$, the equation is a quadratic in $u = 3^x$.

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

Undo the substitution. From $3^x = 1 = 3^0$ we get $x = 0$, and from $3^x = 3 = 3^1$ we get $x = 1$. Both $u$ values are positive, so both give a real $x$.

The solutions are $x = 0$ and $x = 1$.

The arguments require $x + 3 > 0$ and $x - 3 > 0$, so the domain is $x > 3$. Combine with the product rule and convert to exponential form.

$(x + 3)(x - 3) = 4^2 \;\Longrightarrow\; x^2 - 9 = 16 \;\Longrightarrow\; x^2 = 25$

The candidates are $x = 5$ and $x = -5$. Only $x = 5$ lies in the domain, and it checks out, since $\log_4 8 + \log_4 2 = \log_4 16 = 2$.

At $x = -5$ the two arguments are $-2$ and $-8$, both negative, yet their product is $16$, which is exactly what the combined equation wants. That is how the combining step manufactured a root the original equation never had.

The arguments require $x > 0$. Use the quotient rule and convert to exponential form.

$\log_2 \frac{x + 7}{x} = 3 \;\Longrightarrow\; \frac{x + 7}{x} = 2^3 = 8 \;\Longrightarrow\; x + 7 = 8x \;\Longrightarrow\; x = 1$

The value $x = 1$ is positive, so it passes the domain test. Checking, $\log_2 8 - \log_2 1 = 3 - 0 = 3$.

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

$\log_5 x + \log_5(x - 4) = 2 \;\Longrightarrow\; x(x - 4) = 5^2 = 25 \;\Longrightarrow\; x^2 - 4x - 25 = 0$

The quadratic does not factor, so use the quadratic formula.

$x = \frac{4 \pm \sqrt{16 + 100}}{2} = \frac{4 \pm 2\sqrt{29}}{2} = 2 \pm \sqrt{29}$

Since $\sqrt{29} \approx 5.385$, the candidates are about $7.385$ and $-3.385$. Only $2 + \sqrt{29}$ exceeds $4$, so the other candidate is extraneous and the solution is $x = 2 + \sqrt{29}$.

Write every power on base $2$, using $4 = 2^2$ and $8 = 2^3$, then use the product rule for exponents on the left.

$2^x \cdot 2^{2x+2} = 2^{6x-3} \;\Longrightarrow\; 2^{3x+2} = 2^{6x-3}$

Now the bases match, so the exponents must be equal.

$3x + 2 = 6x - 3 \;\Longrightarrow\; 3x = 5 \;\Longrightarrow\; x = \frac{5}{3}$

Convert to exponential form. The statement says the exponent $\frac52$ on the base $x$ produces $32$.

$x^{5/2} = 32 \;\Longrightarrow\; x = 32^{2/5} = \left(2^5\right)^{2/5} = 2^2 = 4$

Raising both sides to the reciprocal power $\frac25$ undoes the $\frac52$. Checking, $4^{5/2} = \left(\sqrt{4}\right)^5 = 2^5 = 32$, and $x = 4$ is a legal base, being positive and not $1$.

Work from the outside in. The outer logarithm says that its argument, which is $\log_2 x$, equals $2^2$.

$\log_2 x = 2^2 = 4 \;\Longrightarrow\; x = 2^4 = 16$

Check both layers: $\log_2 16 = 4$, and $\log_2 4 = 2$. The inner argument $16$ is positive and the outer argument $4$ is positive, so the value is legal.

This is a quadratic in the quantity $\log_3 x$, so substitute $u = \log_3 x$.

$u^2 - 3u + 2 = 0 \;\Longrightarrow\; (u - 1)(u - 2) = 0 \;\Longrightarrow\; u = 1 \ \text{ or } \ u = 2$

Undo the substitution. From $\log_3 x = 1$ we get $x = 3^1 = 3$, and from $\log_3 x = 2$ we get $x = 3^2 = 9$. Both are positive, so both are legal arguments.

The values $1$ and $2$ are the $u$ values (the logarithms), not the $x$ values.

Combine the right side with the product rule, then equate the arguments.

$\log_5(2x - 1) = \log_5(3x) \;\Longrightarrow\; 2x - 1 = 3x \;\Longrightarrow\; x = -1$

Now test the single candidate in the original equation. The right side contains $\log_5 x$, which needs $x > 0$, and $x = -1$ fails that. (The left argument is $2(-1) - 1 = -3$, also negative.)

Every candidate is extraneous, so the equation has no solution. Combining logarithms produced an equation with a root that the original never had.

The right side is the reciprocal of $\frac{8}{27} = \left(\frac23\right)^3$, and a reciprocal is a negative exponent.

$\frac{27}{8} = \left(\frac{2}{3}\right)^{-3} \;\Longrightarrow\; \left(\frac{2}{3}\right)^x = \left(\frac{2}{3}\right)^{-3} \;\Longrightarrow\; x = -3$

The base $\frac23$ is less than $1$, so its powers shrink as the exponent grows. Reaching a value bigger than $1$ therefore takes a negative exponent.

Combine the logarithms with the product rule and convert to exponential form.

$\log_2(xy) = 5 \;\Longrightarrow\; xy = 2^5 = 32$

Now substitute $x = 4y$ into that product.

$4y^2 = 32 \;\Longrightarrow\; y^2 = 8 \;\Longrightarrow\; y = 2\sqrt{2} \ \text{ or } \ y = -2\sqrt{2}$

The original equation contains $\log_2 y$, so $y$ must be positive and the negative root is extraneous. That leaves $y = 2\sqrt{2}$ and

$x = 4y = 8\sqrt{2}.$

Checking, $xy = 8\sqrt2 \cdot 2\sqrt2 = 32$, and $\log_2 32 = 5$.