SOLUTION: log(3)(4x-7)=2 log(4)x+log(4)(x-3)=1 3^x+1=27 4^x-1=64 help me please!

Question 85823: log(3)(4x-7)=2
log(4)x+log(4)(x-3)=1
3^x+1=27
4^x-1=64
help me please!
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Thought to remember: log is another name for exponent.
-----------------
log(3)(4x-7)=2
Here you are told "the log is 2"
So, 4x-7 = 3^2
4x-7=9
4x=16
x=4
-------------
log(4)x+log(4)(x-3)=1
log(4)[x(x-3)]=1
Here you are told "the log is 1"
x(x-3)=4^1
solve for x
===============
3^(x+1)=27
Rewrite as
3^(x+1) = 3^3
x+1=3
x=2
-----------
4^(x-1)=64
Rewrite as
4^(x-1) = 4^3
x-1=3
x=4
------------
Cheers,
Stan H.

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