Question

How do you solve `5 (2^(3x)) = 8`?

Answer 1

`x = \frac{log_2(\frac{8}{5})}{3}`

Explanation:

We must isolate the `x`: first of all, get rid of the `5`, dividing both sides by `5`:

`2^(3x) = \frac{8}{5}`

Now, take logarithm (base `2`) to both sides:

`3x = log_2(\frac{8}{5})`

Finally, divide both sides by `3`:

`x = \frac{log_2(\frac{8}{5})}{3}`

If you prefer, you can use the rule `log(a/b) = log(a)-log(b)` to write

`log_2(\frac{8}{5}) = log_2(8) - log_2(5)=3-log_2(5)`

And the answer becomes

`x = 1-\frac{log_2(5)}{3}`

Answer 2

`x = (log 1.6)/(3log(2)) = (ln 1.6)/(3ln(2)) ~~ .226`

Explanation:

Given: `5(2^(3x)) = 8`

First divide by `5: " "2^(3x) = 8/5 = 1.6`

Log base 2 both sides: `" " log_2 2^(3x) = log_2 1.6`

Use the logarithmic property: `log_b b^x = x`

`3x = log_2 1.6`

`x =( log_2 1.6)/3 = 1/3 ( log_2 1.6)`

Use the change of base formula to convert to either log base 10 or the natural log: `log_b x =( log x)/(log 2) = ( ln x)/(ln 2)`

`x = 1/3 (log 1.6)/(log(2)) = (log 1.6)/(3log(2)) ~~.226`

Answer 3

`x=1-1/3log_2(5)`

Explanation:

You need to work with exponents here.

`5*2^(3x)=8=2^3`
That means `2^(3x)=2^3/5`
Divide with `2^3` on both sides: `2^(3(x-1))=1/5`

Take `log_2` on both sides:

`3(x-1)=-log_2(5)`

or `x=1-1/3log_2(5)`

Check:
`5*2^(3(1-1/3log_2(5))` =`5*2^(3-log_2(5)`=`5*8/5=8`