Question

How do you solve `8^x=1000`?

Answer 1

`x~~3.322`

Explanation:

Convert the exponential form to a logarithmic form `a^x=b-> x=log_ab`

`8^x=1000`

`x=log_8 1000`

You can use the 'change of base law' to calculate it.

`log_a b = (log_c b)/(log_c a)" "` (c is usually 10)

`x = log_10 1000//log_10 8`

`x~~3.322`

Answer 2

`x≈3.322`

Explanation:

Use the `color(blue)"law of logarithms"`

`color(red)(bar(ul(|color(white)(2/2)color(black)(logx^n=nlogx)color(white)(2/2)|)))`
Applies to logarithms to any base.

Take the ln ( natural log) of both sides.

`rArrln8^x=ln1000`

Using the above law.

`rArrxln8=ln1000`

divide both sides by ln8

`(x cancel(ln8))/cancel(ln8)=ln1000/ln8`

`rArrx≈3.322" to 3 decimal places"`