Find the number of irrational solutions of the equation $$3^x8^{\frac{x}{x+1}}=36.$$
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Hi and welcome to the site! Since this is a site that encourages learning, you will get much more help if you show us what you have already done. Could you edit your question with your thoughts and ideas? – 5xum Aug 08 '14 at 06:16
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Take the $\log$ and write your equation $A(x)\log 2+B(x)\log 3=0$. – Kelenner Aug 08 '14 at 06:29
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Write exponents in LHS of ${3}^{x} {8}^{\frac{x}{x+1}} = 36$ in terms of a common base:
$${3}^{x} {8}^{\frac{x}{x+1}} = {e}^{\ln{3^x}} e^{\ln{{8}^{\frac{x}{x+1}}}} = {e}^{x \ln{3}} {e}^{\frac{{x} \ln{8}}{x+1}} = {e}^{x \ln{3}+\frac{x \ln{8}}{x+1}} = 36$$
Taking the natural log of both sides gives $x\ln{3} + \frac{x\ln{8}}{x+1} = \ln{36}$, and writing the LHS as a single fraction gives $$\frac{x\ln{8}+{x}^{2}\ln{3}+x\ln{3}}{x+1} = \ln{36}$$ By clearing fractions and bringing together like terms we get a quadratic in $x$:$$(\ln{3}){x}^{2}+(\ln{3}+\ln{8}-\ln{36})x-\ln{36} = 0$$ And after applying the quadratic formula we get the roots
$$x = \frac{-\ln{3}-\ln{8}+\ln{36}\pm\sqrt{{(\ln{3}+\ln{8}-\ln{36})}^{2}+4\ln{3}\ln{36}}}{2\ln{3}}$$ Which, after drastic simplification, reduce to the trivial root $2$ and the irrational root $-\frac{\ln{2}+\ln{3}}{\ln{3}}$. Since there is one irrational root, your answer is $\boxed{1}$.
I realize there is probably a more efficient way to solve this problem, but I don't feel like thinking of one as of the moment.
mk1123
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