# precal

logs
9^(2x-5)=27(x+4)

and

3log(underscore4)(2x-1)=15

Thank you

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1. you typed 9^(2x-5)=27(x+4)

I assume you meant to say 9^(2x-5)=27^(x+4)

notice both bases are powers of 3, so let's change the bases

(3^2) ^(2x+5) = (3^3)^x+4)

3^(4x+10) = 3^(3x+12)

since the bases are the same, the exponents must be the same.
4x+10 = 3x+12
x = 2

for your second question of 3log(underscore4)(2x-1)=15
divide both sides by 3, then change it to an exponential equation

log(underscore4)(2x-1)=5
2x-1 = 4^5
2x-1 = 1024
x = 1025/2 or 512 1/2 or 512.5

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