Question

How do you solve 16 + 2^x = 2^(x+1) algebraically without using log?

Answer 1

See below. `x = 4`.

Explanation:

Recall that any exponential `a^x` can be written also as, for example, `a*a^(x-1)`, or `a^2*a^(x-2)`, etc. When multiplying two exponential terms that have the same base, the result is that base raised to the sum of the powers of the component parts.

Also recall that `na = sum_1^n a`; that is, multiplying a by `n` is the same as adding a `n` times.
Using that, we can determine...

`16 + 2^x = 2^(x+1) = 2*2^x = 2^x + 2^x -> 16 = 2^x`

Now we look at the factors of 16; all of the prime factors of 16 are 2. Specifically, 16 is equal to `2*2*2*2 = 2^4`.

Thus, `2^4 = 2^x -> x = 4`