SOLUTION: Can you please help me solve the equation {{{ 3^(2x-4) +5=70 }}}? I got as far as {{{ log(3, 2x-4)=65 }}} and now I am stuck.

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: Can you please help me solve the equation {{{ 3^(2x-4) +5=70 }}}? I got as far as {{{ log(3, 2x-4)=65 }}} and now I am stuck.      Log On


   



Question 753700: Can you please help me solve the equation +3%5E%282x-4%29+%2B5=70+?
I got as far as +log%283%2C+2x-4%29=65+ and now I am stuck.

Found 2 solutions by stanbon, solver91311:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
3^(2x-4) +5=70
---
3^(2x-4) = 65
---
Take the log of both sides to get:
(2x-4)log(3) = log(65)
----
2x-4 = log(65)/log(3)
----
2x-4 = 3.7997
2x = 7.7997
x = 3.8998
==============================
Cheers,
Stan H.
=============

Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!


You are stuck because you failed to take the log of the RHS when you took the log of the LHS. Furthermore, you didn't take the base 3 log of the LHS correctly.







Now use



Now use



From here, just solve for remembering that the log of a constant is just another numerical constant.

John

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