SOLUTION: 1. log2(x-3)=3-log2(x-1) (the base is log 2) 2. 6log7(x^2-1)-5log7(x-1) (the base is log 7) For the first problem i need to solve it and for the 2nd problem i need to express

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: 1. log2(x-3)=3-log2(x-1) (the base is log 2) 2. 6log7(x^2-1)-5log7(x-1) (the base is log 7) For the first problem i need to solve it and for the 2nd problem i need to express       Log On


   

Question 928299: 1. log2(x-3)=3-log2(x-1) (the base is log 2)
2. 6log7(x^2-1)-5log7(x-1) (the base is log 7)
For the first problem i need to solve it
and for the 2nd problem i need to express it as a single logarithm.
I have been trying to solve this but I am always off on my answers. I need to know how it is solve step by step. Thank you so much.
Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!
1. solve it
log%282%2C%28x-3%29%29=3-log%282%28x-1%29%29

log%282%2C%28x-3%29%29%2Blog%282%28x-1%29%29=3
log%282%2C%28%28x-3%29%28x-1%29%29%29=3 ...in base 10
log%28%28%28x-3%29%28x-1%29%29%29%2Flog%282%29=3
log%28%28%28x-3%29%28x-1%29%29%29=3log%282%29
log%28%28%28x-3%29%28x-1%29%29%29=log%282%5E3%29
log%28%28%28x-3%29%28x-1%29%29%29=log%288%29 ...if log. same, then
%28x-3%29%28x-1%29=8 is same too; so, solve it for x
x%5E2-x-3x%2B3=8
x%5E2-4x%2B3-8=0
x%5E2-4x-5=0
x%5E2%2Bx-5x-5=0...group
%28x%5E2%2Bx%29-%285x%2B5%29=0
x%28x%2B1%29-5%28x%2B1%29=0
%28x-5%29%28x%2B1%29=0
solutions:
if x-5=0=> highlight%28x=5%29...your answer
if x%2B1=0=> x=-1...reject negative value for log

2. express it as a single logarithm
6log%287%2C%28x%5E2-1%29%29-5log%287%2C%28x-1%29%29

log%287%2C%28x%5E2-1%29%5E6%29-log%287%2C%28x-1%29%5E5%29

log%287%2C%28%28x%5E2-1%29%5E6%2F%28x-1%29%5E5%29%29....since %28x%5E2-1%29%5E6=%28%28x-1%29%28x%2B1%29%29%5E6=%28x-1%29%5E6%28x%2B1%29%5E6, we will have

log%287%2C%28%28%28x-1%29%5E6%28x%2B1%29%5E6%29%2F%28x-1%29%5E5%29%29%29%29



log%287%2C%28%28x-1%29%28x%2B1%29%5E6%29%29