Rewriting in Exponential Form

Solve $\log_2 x = 3$.

Identify the parts: Base $2$, argument $x$, value of the logarithm $3$ = $\log_2 x = 3$

Rewrite in exponential form: $\log_b A = c \iff A = b^c$ = $x = 2^3$

Calculate the power: $2^3 = 2 \cdot 2 \cdot 2$ = $x = 8$

Check: $\log_2 8 = 3$ because $2^3 = 8$ ✓, and the argument $8$ is positive = The solution is valid

Answer: $x = 8$

An Expression Inside the Logarithm

Solve $\log_3(2x + 1) = 2$.

Rewrite in exponential form: The base $3$ is raised to the value $2$: $2x + 1 = 3^2$ = $2x + 1 = 9$

Solve the linear equation: $2x = 9 - 1 = 8$ = $x = 4$

Check the domain: Argument: $2 \cdot 4 + 1 = 9 > 0$ = The argument is positive

Check the equation: $\log_3 9 = 2$ because $3^2 = 9$ ✓ = The solution is valid

Answer: $x = 4$

Logarithms on Both Sides

Solve $\log_5(3x - 2) = \log_5(x + 6)$.

Use the one-to-one property: Both sides are single logarithms with base $5$, so the arguments are equal = $3x - 2 = x + 6$

Solve the linear equation: $3x - x = 6 + 2$, so $2x = 8$ = $x = 4$

Check both arguments: $3 \cdot 4 - 2 = 10 > 0$ and $4 + 6 = 10 > 0$ = Both arguments are positive ✓

Answer: $x = 4$

Condense First: Two Candidates, One Rejected

Solve $\log_2 x + \log_2(x - 2) = 3$.

Condense with the product rule: $\log_2 x + \log_2(x - 2) = \log_2\big(x(x - 2)\big)$ = $\log_2\big(x(x - 2)\big) = 3$

Rewrite in exponential form: $x(x - 2) = 2^3$ = $x^2 - 2x = 8$

Solve the quadratic equation: $x^2 - 2x - 8 = 0$, so $(x - 4)(x + 2) = 0$ = Candidates: $x = 4$ or $x = -2$

Check every candidate in the original equation: $x = 4$: the arguments $4$ and $2$ are positive ✓. $x = -2$: $\log_2(-2)$ is not defined ✗ = $x = -2$ is an extraneous solution

Answer: $x = 4$ (the candidate $x = -2$ is rejected). The second graph shows why: the condensed curve $y = \log_2\big(x(x-2)\big)$ also exists for negative $x$ and meets $y = 3$ twice, but the original curve meets it only at $x = 4$. For $x > 2$ the two curves lie on top of each other; switch them on and off in the legend to compare.

Different Bases: Take the Logarithm of Both Sides

Solve $3^x = 20$. Round to two decimal places.

Look for a common base: $3^2 = 9$ and $3^3 = 27$, so $20$ is not a power of $3$ = The same-base method fails, but $2 < x < 3$

Take the common logarithm of both sides: $\log(3^x) = \log 20$, then the power rule = $x \log 3 = \log 20$

Divide by $\log 3$: $\log 3$ is a number, about $0.4771$ = $x = \frac{\log 20}{\log 3}$ (exact)

Use the calculator: $\frac{1.3010}{0.4771} \approx 2.727$ = $x \approx 2.73$

Check: $3^{2.73} \approx 20.07 \approx 20$ ✓, and $2.73$ lies between $2$ and $3$ = The solution is valid

Answer: $x = \frac{\log 20}{\log 3} \approx 2.73$

Mistake: Keeping both candidates: $\log_2 x + \log_2(x - 2) = 3 \implies x = 4$ or $x = -2$

Why: After condensing to $\log_2\big(x(x - 2)\big)$, negative values of $x$ seem allowed, because the product of two negative numbers is positive. But the original equation contains $\log_2 x$, which is not defined for $x = -2$.

Correct: Check every candidate in the ORIGINAL equation. Only $x = 4$ works; $x = -2$ is extraneous.

Mistake: Dropping the logarithms: $\log_2 x + \log_2(x - 2) = 3 \implies x + (x - 2) = 3$

Why: You may remove logarithms only when each side is a single logarithm with the same base (one-to-one property). Here the left side is a sum of logarithms and the right side is a number.

Correct: Condense first, $\log_2\big(x(x - 2)\big) = 3$, then rewrite in exponential form: $x(x - 2) = 2^3 = 8$.

Mistake: Swapping base and exponent: $\log_3(2x + 1) = 2 \implies 2x + 1 = 2^3$

Why: In $\log_b A = c$ the base $b$ is raised to the power $c$. The base of the logarithm stays the base of the power.

Correct: $2x + 1 = 3^2 = 9$, so $x = 4$. Check: $\log_3 9 = 2$ ✓

Mistake: Dividing instead of taking logarithms: $3^x = 20 \implies x = \frac{20}{3} \approx 6.67$

Why: $3^x$ is a power, not the product $3 \cdot x$. A quick check shows the error: $3^{6.67}$ is about $1500$, far more than $20$.

Correct: $x = \frac{\log 20}{\log 3} \approx 2.73$. Note that $\frac{\log 20}{\log 3}$ is NOT the same as $\log \frac{20}{3}$.

Doubling Your Savings

Money that grows by a fixed percentage each year follows $A = P \cdot (1 + r)^t$. Solving for the exponent $t$ with logarithms tells you when a savings goal is reached.

At 6% interest a year, money doubles when $1.06^t = 2$, so $t = \frac{\ln 2}{\ln 1.06} \approx 11.9$ years.

Loudness in Decibels

The sound level in decibels is $L = 10 \log r$, where $r = \frac{I}{I_0}$ compares the intensity of a sound with the quietest sound people can hear.

Normal conversation is about 60 dB: $10 \log r = 60$ gives $\log r = 6$, so $r = 10^6$, one million times the threshold of hearing.

Growing Bacteria

A colony that doubles every hour follows $N = N_0 \cdot 2^t$. When the target is not a power of 2, logarithms give the exact time.

A colony of 200 cells reaches 1000 cells when $2^t = 5$, so $t = \frac{\ln 5}{\ln 2} \approx 2.32$ hours.

Q: Why do extraneous solutions appear at all?

A: Condensing changes the domain. $\log_2 x + \log_2(x - 2)$ needs $x > 2$, but $\log_2\big(x(x - 2)\big)$ only needs $x(x - 2) > 0$, which is also true for $x < 0$. The condensed equation can therefore have solutions the original one does not have. That is why you always check in the original equation.

Q: Is a negative solution always extraneous?

A: No. What matters is the sign of each argument, not the sign of $x$. In $\log_2(x + 5) = 1$ we get $x + 5 = 2$, so $x = -3$, and the argument $-3 + 5 = 2$ is positive. The solution $x = -3$ is valid.

Q: Should I use log or ln to solve $3^x = 20$?

A: Either one. $\frac{\log 20}{\log 3}$ and $\frac{\ln 20}{\ln 3}$ give the same number, about $2.73$. This is exactly the change of base formula: both are equal to $\log_3 20$.

Q: Can a logarithmic equation have no solution?

A: Yes. For $\log_2(x - 4) = \log_2(2x - 1)$ the arguments give $x - 4 = 2x - 1$, so $x = -3$. But then both arguments are $-7$, which is not allowed. The only candidate is rejected, so the equation has no solution.