SOLUTION: Solve and check a) 1og(base 7)(x+4) + log(base 7)(x-2) = 1 b) log(base 2)(2m+4) - log(base 2)(m-1) = 3 Can you please help me out? Thanks so much in advance Can you also show

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: Solve and check a) 1og(base 7)(x+4) + log(base 7)(x-2) = 1 b) log(base 2)(2m+4) - log(base 2)(m-1) = 3 Can you please help me out? Thanks so much in advance Can you also show      Log On


   

Question 747370: Solve and check
a) 1og(base 7)(x+4) + log(base 7)(x-2) = 1
b) log(base 2)(2m+4) - log(base 2)(m-1) = 3
Can you please help me out? Thanks so much in advance
Can you also show me the steps it would really help me understand:)
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

log%287%2C%28x%2B4%29+%29%2B+log%287%2C%28x-2%29%29+=+1....in base 10 we will have

log%28%28x%2B4%29%29%2Flog%287%29%2B+log%28%28x-2%29%29%2Flog%287%29+=+1

%28log%28%28x%2B4%29%29%2B+log%28%28x-2%29%29%29%2Flog%287%29+=+1
log%28%28x%2B4%29%28x-2%29%29+=+1%2Alog%287%29
log%28%28x%2B4%29%28x-2%29%29+=+log%287%29
%28x%2B4%29%28x-2%29+=+7
x%5E2-2x%2B4x-8+=+7
x%5E2%2B2x-8-7=0
x%5E2%2B2x-15=0....write 2x as 5x-3x
x%5E2%2B5x-3x-15=0....group
%28x%5E2%2B5x%29-%283x%2B15%29=0
x%28x%2B5%29-3%28x%2B5%29=0
%28x-3%29%28x%2B5%29=0
solutions:
x-3=0 => x=3
x%2B5=0 => x=-5....we do not need this solution since we deal with a logarithm

so, solution to your problem is: x=3

b.
log%282%2C%282m%2B4%29%29+-+log%282%2C%28m-1%29%29+=+3
log%28%282m%2B4%29%29%2Flog%282%29+-+log%28%28m-1%29%29%2Flog%282%29+=+3

%28log%28%282m%2B4%29%29+-+log%28%28m-1%29%29%29%2Flog%282%29+=+3
log%28%282m%2B4%29%29+-+log%28%28m-1%29%29+=+3log%282%29
log%28%282m%2B4%29%2F%28m-1%29%29+=+log%282%5E3%29
log%28%282m%2B4%29%2F%28m-1%29%29+=+log%288%29
%282m%2B4%29%2F%28m-1%29+=+8
2m%2B4+=+8%28m-1%29
2m%2B4+=+8m-8
8%2B4+=+8m-2m
12+=+6m
12%2F6=m
m=2