Posted by **Anonymous** on Friday, September 29, 2006 at 6:05pm.

here is the question:

log5(x-4)= log7x solve for x.

These are just base 10 logs.

log100 = 2

This equation has the same format as log 40 = log (2x20)

Since both sides have log base 10, you divide by log base 10 and end up with 5(x-4) = 7x, so 5x -20 =7x x=-10

One doesn't

*exactly * "divide by base 10".

log5(x-4)= log7x solve for x.

take each side as exponents to the base 10, or

10^(log5(x-4))= 10^(log7x)

then

5(x-4) = 7x

and x= 10

check...

log (5(-14)= log -70

log (-70)=log(-70) and it checks

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