Question 742189: Solve and check
a) 2log(x-1) = 2+ log 100
Found 3 solutions by josgarithmetic, lwsshak3, MathTherapy:
Answer by josgarithmetic(39838)   (Show Source):
Answer by lwsshak3(11628) (Show Source):
You can put this solution on YOUR website! Solve and check
a) 2log(x-1) = 2+ log 100
2log(x-1) = 2+2
2log(x-1) = 4
log(x-1) =2=log100
x-1=100)
x=101
check:
2log(x-1)=2log(100)=2*2=4
2+log 100=2+2=4
Answer by MathTherapy(10858)  (Show Source):
You can put this solution on YOUR website!
Solve and check
a) 2log(x-1) = 2+ log 100
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The response from @josgarithmetic(39831), while correct, is confusing, in this author's opinion. Whenever one sees a log argument without a base,
it's widely known that that base is 10, since that's the base/number system we all work in. This is just the same as x. It's known that this is actually
1x, but it's not written like that, just x. Also, x is actually  , but again, it's not written as such, just x. So, why confuse a person by working the
problem without a base, just because a base wasn't stated? That base, as stated before, is OBVIOUSLY no other than 10.
2 log(x - 1) = 2 + log (100)
2 log(x - 1) = 2 + 2 ---- log (100) = 2
2 log(x - 1) = 4
 ----- Dividing each side by 2
log (x - 1) = 2
 ---- Converting to EXPONENTIAL form
x - 1 = 100
x = 100 + 1 = 101
You can do the CHECK!!
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