Question 742189
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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 {{{x^1}}}, 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
{{{(2(log((x - 1))))/2 = 4/2}}} ----- Dividing each side by 2 
            log (x - 1) = 2
                     {{{x - 1 = 10^2}}} ---- Converting to EXPONENTIAL form
                     x - 1 = 100
                           x = 100 + 1 = 101

You can do the CHECK!!</font></font></font></b></pre>