Need help solving exponential equation! Must solve for x. The equation is 5^x = 25 + 5^(x-3)?

Question

Any help is appreciated ASAP! All of the methods i used so far have resulted in x getting canceled out. Thank you!

Answer 1

First rewrite 25 as 5^2
Then divide everything by 5^x. Dividing means subtract the exponents so you get

1 = 5^(2-x) + 5^-3

subtract 5^-3 from both sides getting

124/125 = 5^(2-x)

take log of both sides
log(124/125) = (2-x) log 5

divide both sides by log 5, then subtract 2 from both sides then divide by -1 and you'll get x.

I left out some stuff so if this isn't enough send an IM

Answer 2

5^x = 25 + 5^(x-3) = 5^2 + 5^(x-3)

Therefore,

5^(x - x) = 5^(2 - x) + 5^(x - 3 - x)

1 = 5^(2 - x) + 5^(-3)

5^(2-x) = 1 - 5^(-3)

Now take ln() of both sides and use ln(a^x) = x ln( a )
(ln() is the natural log)

(2 - x) ln( 5 ) = ln (1 - 5^(-3))

x = 2 - [ ln(1 - 5^(-3)) / ln(5) ]

x is approx 2.005