Exponential equations

For $3x = 12$ you divide by 3. For $x^3 = 8$ you take the cube root. But what do you do with $3^x = 81$? Here the $x$ is not the base but the exponent, and none of the familiar inverse operations reaches it. The logarithm exists exactly for this situation.

When guessing is enough

Sometimes you need no calculator at all. You solve $3^x = 81$ by running through the powers of 3: 3, 9, 27, 81. That is the fourth step, so $x = 4$. Always try this route first, because it is faster and gives an exact number instead of a rounded one.

The logarithm as a tool

For $2^x = 50$ trying gets you nowhere: $2^5 = 32$ is too small, $2^6 = 64$ too large, the solution lies somewhere in between. Now you apply the logarithm to both sides. The decisive step is the rule $\log(a^x) = x \cdot \log(a)$, because it slides the exponent down and turns it into an ordinary factor.

From $2^x = 50$ you get $\log(2^x) = \log(50)$, so $x \cdot \log(2) = \log(50)$. And now it is a perfectly ordinary equation: you divide by $\log(2)$ and get $x = \frac{\log(50)}{\log(2)} \approx 5.64$. That matches the estimate from before, because 5.64 lies between 5 and 6.

Isolate it first

In real tasks the power rarely stands alone. For $5 \cdot 2^x = 320$ you must not apply the logarithm straight away, because it acts on the whole product and not just on the power. First you divide by 5 and get $2^x = 64$, and then you are back to a task you can even do in your head: $x = 6$. So the order is the same as for any equation: clear everything in the way first, then apply the inverse operation.