(1) -4 + 7log(x) = 10
7log(x) = 14
log(x) = 2
x = 10^2 = 100
(2) log(x+3) + log(x-2) = log 14
recall some laws of exponents. when we add two log terms with same base, we can combine them by multiplying the terms inside the log. thus, we can rewrite this as,
log[(x+3)(x-2)] = log 14
equating the terms inside,
(x+3)(x-2) = 14
x^2 + x - 6 = 14
x^2 + x - 20 = 0
since it's factorable,
(x+5)(x-4) = 0
x = -5 and x = 4
but note that x = -5 is extraneous since if this substituted back to original,
log(-5+3) + log(-5-2)
log(-2) + log(-7)
..and log of any negative number is undefined, thus
x = 4 only.
hope this helps~ :)
solve
-4+7logx=10
&
log(x+3)+log(x-2)=log14
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