Solving an Exponential Equation by Taking a Logarithm

Ordinary algebra undoes operations by reversing them in the order they were applied, but an exponent isn't attached by addition or multiplication, so adding or dividing never reaches it. Dividing both sides of 3^x = 200 by 3 does not give x = 200/3 - it just peels off one factor of 3, leaving 3^x-1 on the left. The exponent is still stuck upstairs.

A logarithm is the one operation built to undo exponents, because log_b(b^x) = x. Take log (or ln) of both sides of 3^x = 200: ln(3^x) = ln(200). The power law of logs says ln(3^x) = x ln(3), so the exponent drops down and becomes an ordinary multiplier: x ln(3) = ln(200), giving x = ln(200)/ln(3), about 4.823.

The same move answers growth and decay questions phrased in years. For 1200 dollars growing at 6 percent a year to reach 2000, write 1200(1.06)^t = 2000, divide to get (1.06)^t = 1.667, then take logs: t = ln(1.667)/ln(1.06), about 8.77 years. When the exponent is something like 2x + 1 instead of a bare x, the log still drags the whole exponent down at once - ln(b^2x+1) = (2x+1)ln(b) - and then normal algebra (subtract 1, divide by 2) finishes isolating x, linking this topic straight back to the log laws and inverse-function ideas it depends on.