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My try:

I tried plotting the graph(manually) of LHS and RHS but it does not give any information whether they will intersect or not.If they will intersect then there will be 1 solution and if they do not intersect then there will be no solution. But the main problem I found was to check if they intersect or not.

Please do not use the aid of Wolfram Alpha, graphing calculator while answering this question.

user2369284
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1 Answers1

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Divide both sides by $6^x$: $$\tag1\left(\frac12\right)^x+\left(\frac23\right)^x+\left(\frac56\right)^x=1.$$ Note that $f(x)=q^x$ with $0<q<1$ has the property that $f(0)=1$, $f$ is continuous and strictly decreasing on $[0,\infty)$ and $f(x)\to 0$ as $x\to\infty$. Therefore the left hand side in $(1)$ is $3$ at $x=0$, is continuous and structliy decreasing on $[0,\infty)$ and converges to $0$ as $x\to\infty$. By the IVT, there exists at least one $x$ and by monotonicity at most one $x$ such that the LHS equals $1$.

Hagen von Eitzen
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    Great answer!!! – user2369284 Apr 13 '14 at 11:01
  • More over, since $\left(\frac12\right)^3+\left(\frac23\right)^3+\left(\frac56\right)^3=1$, we can state there are no solutions $>3$...and since $f(x) = \left(\frac12\right)^3+\left(\frac23\right)^3+\left(\frac56\right)^3$ is striclty decreasing on $[0,+\infty)$, there are no solutions $<3$. So there is only one solution ($x=3$). – sirfoga Apr 13 '14 at 11:10