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

Question

jhonyy9 · Answer

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

kj4uts · Answer

so is that 27^3?

anonymous · Answer

$$
3^1=3\3^2=9\3^3=27
$$Therefore $3^{-4+3x} = 3^3\implies -4+3x=3$.

kj4uts · Answer

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

anonymous · Answer

Yes

kj4uts · Answer

3x=7 and 7 / 3 = 2.33

kj4uts · Answer

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

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

You will have to use a calculator for this.

kj4uts · Answer

log base 5 (392) = 3.71

anonymous · Answer

Then solve for $x$.

kj4uts · Answer

oh I see plus 4 = 7.71 / 2 = 3.85

kj4uts · Answer

thank you