Question

How do you solve `4^x = 1/2`?

Answer 1

`x=-1/2`

Explanation:

Take the natural logarithm of both sides of the equation

`ln|4^x|=ln|1/2|`

Rewrite using properties of exponents

`xln|4|=ln|1/2|`

`xln|2^2|=ln|1|-ln|2|`

`2xln|2|=0-ln|2|`

`x=(-ln|2|)/(2ln|2|)`

`x=-1/2`

Answer 2

Alternate answer:

Explanation:

Rewriting the exponents in a common base:

Don't forget that `1/(a^n) = a^-n`. Thus, `1/2 = 2^-1`.

`(2^2)^x = 2^-1`

Using the exponent property `(a^n)^m = a^(n xx m)`, we get the following:

`2^(2x) = 2^-1`

We can eliminate the bases now and solve the simple linear equation.

`2x = -1`

`x = -1/2`

Here are a few helpful exponent rules to know when working with harder problems:

`•a^n xx a^m = a^(n + m)`

`•a^n/a^m = a^(n - m)`

`•a^(n/m) = root(m)(a^n)`

Practice exercises:

1. Solve for x.

`3^(2x + 1) xx 9^(x - 3) = 27^(4x - 5)`

`(4^(2x + 5))/(8^(3x - 2)) = 16^(2x)`

Challenge Problem:

Solve for x in the equation `sqrt(12) xx root(x)(12) = root(15)(12)`

Good luck!