Question

How do you solve 3^(-4+3x)=27 and 5^(2x-4)=392? Round to two decimal places. Thank you!

Answer 1

like a first step rewrite 27 in an exponential form of 3

Answer 2

so is that 27^3?

Answer 3

\[
3^1=3\\3^2=9\\3^3=27
\]Therefore \(3^{-4+3x} = 3^3\implies -4+3x=3\).

Answer 4

Oh I see so do I now add -4 + 4 to the 3 on the other side?

Answer 5

Yes

Answer 6

3x=7 and 7 / 3 = 2.33

Answer 7

I am going to try the second one but how do you determine the exponential easily because the second one is 392

Answer 8

You have to use this property: \[
b =a^{\log_a(b)}
\]So: \[
392 = 5^{\log_5(392)}
\]

Answer 9

\[
5^{2x-4}=5^{\log_5(392)} \implies 2x-4 =\log_5(392)
\]

Answer 10

You will have to use a calculator for this.

Answer 11

log base 5 (392) = 3.71

Answer 12

Then solve for \(x\).

Answer 13

oh I see plus 4 = 7.71 / 2 = 3.85

Answer 14

thank you