Posted by **Jenny** on Monday, December 6, 2010 at 10:33pm.

Suppose that log2 = a and log3 = b

Solve for x in terms of a and b:

6^x = 10/3 - 6^-x

- calculus-watch parentheses -
**MathMate**, Monday, December 6, 2010 at 11:32pm
Parentheses are not needed for fractions in type-set texts because it is graphically clear. When transferred to a one-line expression, parentheses are required around the numerator and denominator.

I assume the question is as follows, which is different from what you posted because of the missing parentheses.

6^x=10/(3-6^(-x)) ....(1)

From log(2)=a, and log(3)=b,

we deduce that

e^{a}=2, and

e^{b}=3.

cross multiply (1), assume 3≠6^(-x)

6^x(3-6^(-x))=10

3*6^x - 6^0 = 10 .... (6^0 = 1)

3*6^x = 11

6^x = 11/3

take log on both sides

x(log(2)+log(3) = log(11)-log(3)

x = (log(11)-b)/(a+b)

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