calculus
posted by Jenny .
Suppose that log2 = a and log3 = b
Solve for x in terms of a and b:
6^x = 10/3  6^x

Parentheses are not needed for fractions in typeset texts because it is graphically clear. When transferred to a oneline 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/(36^(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(36^(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)