Solving equations with indices

An SAT Math micro-topic under Radicals and rational exponents (Advanced Math). Free to read — no account needed.

An equation with indices (exponents) is easiest to solve when both sides share the same base.
If

b^{\text{something}} = b^{\text{something else}}

and the base

b

is the same, then the two exponents must be equal.

So you just set the exponents equal and solve the simpler equation.
For

5^{2x} = 5^8

, the base is

5

on both sides, so

2x = 8

, which gives

x = 4

.
The exponents behave like an ordinary equation once the bases match.

Often the two sides look different but can be made to share a base.
Since

4 = 2^2

, the equation

4^x = 2^6

becomes

(2^2)^x = 2^6

, that is

2^{2x} = 2^6

.
Now the bases match, so

2x = 6

and

x = 3

The same idea works with inequalities as long as the base is greater than

1

.
For

2^{3n} > 2^{13}

the bases match, so

3n > 13

, giving

n > \frac{13}{3}

.
The smallest whole number that satisfies this is

n = 5

Worked examples

Solve

2^{2x} = 2^{10}

.
The base is

2

on both sides, so the exponents are equal:

2x = 10

.
Dividing by

2

x = 5

Solve

9^x = 3^8

.
Rewrite the left side with base

3

: since

9 = 3^2

9^x = (3^2)^x = 3^{2x}

.
Now

3^{2x} = 3^8

, so

2x = 8

and

x = 4

7^{m+1} = 7^5

, what is

m

?
The base is

7

on both sides, so the exponents are equal:

m + 1 = 5

.
Subtracting

1

m = 4

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