# calculus

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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 -

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
ea=2, and
eb=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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