PLEASE HELP WITH ALGERBRA 2!!

Hii!! Can anyone please help me with the problem?? I have no clue how to solve it :/

3^x=8^(x+9)

Thank you so much! I PROMISE I'll choose best answer!

3^x=8^(x+9)
take logs of both sides
x log 3 = (x+9) log 8
x log 3 = x log 8 + 9 log 8
x log 3 - x log 8 = 9 log 8
x ( log 3 - log 8) = 9 log 8
x = 9 log 8 / ( log 3 - log 8)

3^x = 8^(x + 9)
Real solution:
x
= -[27 log (2)] / [3 log (2) - log (3)]
= [27 log (2)] / [log (3) - 3 log (2)]
= -19.080766...

3^x=8^(x+9)

take either log or ln of each side

ln(3^x)=ln(8^(x+9))

remember that ln(x^m)=m*lnx

so:

xln3=(x+9)ln8
xln3=xln8+9ln8
xln3-xln8=9ln8
x(ln3-ln8)=9ln8
x=9ln8/(ln3-ln8)

and there you go :D
hope it helps

Since you cannot rewrite both sides using the same base, it is necessary to use logs.

log(3^x) = log(8^(x + 9))
x log 3 = (x + 9) log 8 = x log 8 + 9 log 8
x log 3 - x log 8 = 9 log 8
x(log 3 - log 8) = 9 log 8
x = (9 log 8)/(log 3 - log 8)

next step:

xlog3=(x+9)log8
x(log3-log8)=9log8
x=(9log8)/(log3-log9)
x= -9log8/log3

note the negative out front

xlog3 = (x+9)log8=xlog8+9log8
x(log3-log8) = 9log8
x = -19.08