AlgebraDifficulty 25-8024 tagged problems

Logarithms

Logarithms is the habit of translating exponent relationships into additive structure for equations and contest estimates. In contest math, that habit turns a crowded setup into a relation the student can test, bound, count, or compute. MathGrit teaches it as a recognizable signal, a deliberate move, and a final translation back to the original question.

Practice this technique Match my level

The Key Result

$$\log_b(xy) = \log_b x + \log_b y,\qquad \log_b\!\left(x^k\right) = k\log_b x,\qquad \log_b x = \frac{\log_c x}{\log_c b}.$$

Logarithms turn multiplication into addition and exponents into multipliers, which is exactly what you need to solve equations where the unknown sits in an exponent.

Why It Works

1
By definition $\log_b x$ is the exponent to which $b$ must be raised to give $x$: if $x = b^u$ and $y = b^v$, then $\log_b x = u$ and $\log_b y = v$.
2
Then $xy = b^u b^v = b^{u+v}$, so $\log_b(xy) = u + v = \log_b x + \log_b y$.
3
Also $x^k = (b^u)^k = b^{ku}$, so $\log_b(x^k) = ku = k\log_b x$.
4
Change of base: writing $x = b^{\log_b x}$ and taking $\log_c$ of both sides gives $\log_c x = (\log_b x)(\log_c b)$, hence $\log_b x = \frac{\log_c x}{\log_c b}$.

Worked Examples

Example 1

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

1
Combine the logs: $\log_2\big(x(x-2)\big) = 3$.
2
Rewrite in exponential form: $x(x - 2) = 2^3 = 8$, so $x^2 - 2x - 8 = 0$.
3
Factor: $(x - 4)(x + 2) = 0$, giving $x = 4$ or $x = -2$; reject $x = -2$ since the logs require $x > 2$.

Answer:

$x = 4$.

Warm-up

Evaluate $\log_2 80 - \log_2 5$ exactly.

1
Combine using the quotient law (a consequence of the product law): $\log_2 80 - \log_2 5 = \log_2\!\dfrac{80}{5}$.
2
Simplify the argument: $\dfrac{80}{5} = 16$.
3
Evaluate: $\log_2 16 = 4$ since $2^4 = 16$.

Answer:

$4$.

Contest level

Solve $\log_3 x + \log_9 x = 3$ for real $x$.

1
Convert the second log to base $3$ via change of base: $\log_9 x = \dfrac{\log_3 x}{\log_3 9} = \dfrac{\log_3 x}{2}$.
2
Let $t = \log_3 x$: the equation becomes $t + \dfrac{t}{2} = 3$, i.e. $\dfrac{3t}{2} = 3$, so $t = 2$.
3
Recover $x$: $\log_3 x = 2 \Rightarrow x = 3^2 = 9$. (Check: $\log_3 9 + \log_9 9 = 2 + 1 = 3$.)

Answer:

$x = 9$.

Olympiad / Challenge

Find all real $x > 0$ satisfying $x^{\log_{10} x} = 100x$.

1
Take $\log_{10}$ of both sides; the power rule moves the exponent out front: $(\log_{10} x)(\log_{10} x) = \log_{10}(100x) = 2 + \log_{10} x$.
2
Let $u = \log_{10} x$: this is $u^2 = 2 + u$, i.e. $u^2 - u - 2 = 0$, which factors as $(u - 2)(u + 1) = 0$.
3
So $u = 2$ or $u = -1$, giving $x = 10^2 = 100$ or $x = 10^{-1} = \tfrac{1}{10}$. (Check $x = \tfrac{1}{10}$: LHS $= (10^{-1})^{-1} = 10$, RHS $= 100\cdot\tfrac{1}{10} = 10$.)

Answer:

$x = 100$ or $x = \dfrac{1}{10}$.

Going Deeper

Generalization: $\log_b$ is the inverse of $x \mapsto b^x$, so every exponent law has a logarithmic mirror image — products become sums, powers become products, and roots become divisions; change of base $\log_b x = \frac{\ln x}{\ln b}$ shows all logarithms are scalar multiples of one another.
Where it appears: AMC equations with the unknown trapped in an exponent, AIME problems chaining several log laws (often after substituting $u = \log x$ to linearize), and estimating the number of digits of a huge number via $\lfloor \log_{10} N\rfloor + 1$.
Pitfall: $\log$ has DOMAIN $x > 0$, so after solving you must discard roots that make any argument $\le 0$ — the worked example's $x = -2$ is exactly such an extraneous solution. Also $\log_b(x + y) \ne \log_b x + \log_b y$: the product law applies to a product inside the log, never to a sum.

Spot the Signal

Look for problems where the key step is translating exponent relationships into additive structure for equations and contest estimates.
You can describe the hard part as translating exponent relationships into additive structure for equations and contest estimates, but a direct attack starts producing clutter.
The problem rewards preserving structure instead of expanding, listing, or guessing too early.

Learn the Move

Start by identify the expression or equation that calls for logarithms, then rewrite the givens around it.
Name the relation that makes Logarithms legal before doing computation.
Use the new relation to replace the messiest part of the problem with a cleaner one.
Translate the result back to the quantity the problem actually asks for.

Avoid These Traps

Do not use Logarithms just because the surface looks familiar; verify the required condition first.
Applying Logarithms because it sounds relevant, without checking the trigger first.
Stopping after spotting the technique instead of finishing the calculation or proof.

MathGrit Coach Note

Let logarithms reveal the algebraic structure; then compute only what remains.

Try it on:

Practice a contest problem where the key step is translating exponent relationships into additive structure for equations and contest estimates.

Back to techniques

AlgebraDifficulty 25-8024 tagged problems

Logarithms

Practice this technique Match my level

The Key Result

$$\log_b(xy) = \log_b x + \log_b y,\qquad \log_b\!\left(x^k\right) = k\log_b x,\qquad \log_b x = \frac{\log_c x}{\log_c b}.$$

Logarithms turn multiplication into addition and exponents into multipliers, which is exactly what you need to solve equations where the unknown sits in an exponent.

Why It Works

1
By definition $\log_b x$ is the exponent to which $b$ must be raised to give $x$: if $x = b^u$ and $y = b^v$, then $\log_b x = u$ and $\log_b y = v$.
2
Then $xy = b^u b^v = b^{u+v}$, so $\log_b(xy) = u + v = \log_b x + \log_b y$.
3
Also $x^k = (b^u)^k = b^{ku}$, so $\log_b(x^k) = ku = k\log_b x$.
4
Change of base: writing $x = b^{\log_b x}$ and taking $\log_c$ of both sides gives $\log_c x = (\log_b x)(\log_c b)$, hence $\log_b x = \frac{\log_c x}{\log_c b}$.

Worked Examples

Example 1

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

1
Combine the logs: $\log_2\big(x(x-2)\big) = 3$.
2
Rewrite in exponential form: $x(x - 2) = 2^3 = 8$, so $x^2 - 2x - 8 = 0$.
3
Factor: $(x - 4)(x + 2) = 0$, giving $x = 4$ or $x = -2$; reject $x = -2$ since the logs require $x > 2$.

Answer:

$x = 4$.

Warm-up

Evaluate $\log_2 80 - \log_2 5$ exactly.

1
Combine using the quotient law (a consequence of the product law): $\log_2 80 - \log_2 5 = \log_2\!\dfrac{80}{5}$.
2
Simplify the argument: $\dfrac{80}{5} = 16$.
3
Evaluate: $\log_2 16 = 4$ since $2^4 = 16$.

Answer:

$4$.

Contest level

Solve $\log_3 x + \log_9 x = 3$ for real $x$.

1
Convert the second log to base $3$ via change of base: $\log_9 x = \dfrac{\log_3 x}{\log_3 9} = \dfrac{\log_3 x}{2}$.
2
Let $t = \log_3 x$: the equation becomes $t + \dfrac{t}{2} = 3$, i.e. $\dfrac{3t}{2} = 3$, so $t = 2$.
3
Recover $x$: $\log_3 x = 2 \Rightarrow x = 3^2 = 9$. (Check: $\log_3 9 + \log_9 9 = 2 + 1 = 3$.)

Answer:

$x = 9$.

Olympiad / Challenge

Find all real $x > 0$ satisfying $x^{\log_{10} x} = 100x$.

1
Take $\log_{10}$ of both sides; the power rule moves the exponent out front: $(\log_{10} x)(\log_{10} x) = \log_{10}(100x) = 2 + \log_{10} x$.
2
Let $u = \log_{10} x$: this is $u^2 = 2 + u$, i.e. $u^2 - u - 2 = 0$, which factors as $(u - 2)(u + 1) = 0$.
3
So $u = 2$ or $u = -1$, giving $x = 10^2 = 100$ or $x = 10^{-1} = \tfrac{1}{10}$. (Check $x = \tfrac{1}{10}$: LHS $= (10^{-1})^{-1} = 10$, RHS $= 100\cdot\tfrac{1}{10} = 10$.)

Answer:

$x = 100$ or $x = \dfrac{1}{10}$.

Going Deeper

Generalization: $\log_b$ is the inverse of $x \mapsto b^x$, so every exponent law has a logarithmic mirror image — products become sums, powers become products, and roots become divisions; change of base $\log_b x = \frac{\ln x}{\ln b}$ shows all logarithms are scalar multiples of one another.
Where it appears: AMC equations with the unknown trapped in an exponent, AIME problems chaining several log laws (often after substituting $u = \log x$ to linearize), and estimating the number of digits of a huge number via $\lfloor \log_{10} N\rfloor + 1$.
Pitfall: $\log$ has DOMAIN $x > 0$, so after solving you must discard roots that make any argument $\le 0$ — the worked example's $x = -2$ is exactly such an extraneous solution. Also $\log_b(x + y) \ne \log_b x + \log_b y$: the product law applies to a product inside the log, never to a sum.

Spot the Signal

Look for problems where the key step is translating exponent relationships into additive structure for equations and contest estimates.
You can describe the hard part as translating exponent relationships into additive structure for equations and contest estimates, but a direct attack starts producing clutter.
The problem rewards preserving structure instead of expanding, listing, or guessing too early.

Learn the Move

Start by identify the expression or equation that calls for logarithms, then rewrite the givens around it.
Name the relation that makes Logarithms legal before doing computation.
Use the new relation to replace the messiest part of the problem with a cleaner one.
Translate the result back to the quantity the problem actually asks for.

Avoid These Traps

Do not use Logarithms just because the surface looks familiar; verify the required condition first.
Applying Logarithms because it sounds relevant, without checking the trigger first.
Stopping after spotting the technique instead of finishing the calculation or proof.

MathGrit Coach Note

Let logarithms reveal the algebraic structure; then compute only what remains.

Try it on:

Practice a contest problem where the key step is translating exponent relationships into additive structure for equations and contest estimates.