Exponential and log equations

An exponential equation holds its unknown hostage in an exponent: $3^x = 20$. No amount of ordinary arithmetic reaches x up there; the key is a logarithm, because a log is an exponent. Either write the answer directly, $x = \log_3 20$, or take a common log of both sides and use the power property: $x\log 3 = \log 20$, so $x = \log 20 / \log 3 \approx 2.73$.

A logarithmic equation hides the unknown inside a log: $\log_3(x) = 4$. The key is exponentiation, rewriting with the definition: $x = 3^4 = 81$. In both directions you are using the same fact, that $\log_b$ and $b^{\,x}$ undo each other.

Sanity-check every answer against the bracketing powers: for $3^x = 20$, since $3^2 = 9$ and $3^3 = 27$ bracket 20, x must sit between 2 and 3. An answer like 6.7 fails that smell test instantly.

When an equation carries two or more logs, condense them into one before exponentiating, using only the three real properties:

1. $\log_2(x) + \log_2(x-6) = 4$ condenses by the product rule to $\log_2\big(x(x-6)\big) = 4$.
2. Convert with the definition: $x(x-6) = 2^4 = 16$.
3. Solve the resulting polynomial: $x^2 - 6x - 16 = 0$, so $(x-8)(x+2) = 0$, candidates $x = 8$ and $x = -2$.

The wrong turn happens at step 1. Splitting or merging with the fake sum property, writing $\log_2(x) + \log_2(x-6)$ as $\log_2(2x - 6)$, produces a clean-looking equation whose answers have nothing to do with the original. Condensing must run through products, quotients, and powers only.

Squaring, condensing, and clearing logs can all enlarge the solution set: the polynomial you end up solving remembers less than the equation you started with. Its roots are candidates, not answers.

Finish the example: candidates 8 and $-2$. Substitute each into the ORIGINAL equation. $x = 8$: $\log_2 8 + \log_2 2 = 3 + 1 = 4$. Keep. $x = -2$: the very first term, $\log_2(-2)$, does not exist. Discard. The solution is $x = 8$ alone, and writing "$x = 8$ or $x = -2$" is a wrong answer, not a generous one.

The subtle version: candidates can satisfy the CONDENSED equation while failing the original, because a product of two negatives is positive. $(-2)(-2-6) = 16$ checks in $x(x-6)=16$, yet both logs of the original are undefined at $x=-2$. That is why the check must use the original equation, log by log, not the condensed one.

Log inequalities carry the same trap in inequality clothing. Solving $\log_2 x < 3$ by converting gives $x < 8$, but the log also demands $x > 0$; the answer is $0 < x < 8$. The unstated left wall is the domain, and dropping it hands back a solution set full of numbers the function cannot even evaluate.

Finally, form. The exact solution of $10^x = 7$ is $x = \log 7$; the decimal 0.845 is an approximation. Leave answers in log form unless a decimal is requested, and when you do approximate, keep the equals signs honest: $x = \log 7 \approx 0.845$.