I am stuck on the following problem which one of my friends gave me:

Solve : $\,8^x=6x$.


We see that $$8^x=6x \implies 2^{3x}=6x.$$ Now I am not sure how to proceed further. Taking logarithm on both sides of the equation does not help much. Clearly, $x=\frac 1 3$ satisfies the given equation but I am not sure how to get it.

Can someone help? Thanks and regards to all.


For simplicity let $3x=u$ so you are really looking at $2^u = 2u$. One is exponential, the other is linear and given the nature of their derivatives, they intersect at most twice. In this case, they intersect twice: at $u=1$ and $u=2$. These are the only solutions. So for your original question that means $x = 1/3$ or $x=2/3$.

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    $\begingroup$ @learner The function $2^u$ has a strictly positive second derivative. Which means it will get steeper and steeper. It can intersect a linear function twice: once being less steep, and once being steeper. After that, $2^u$ will never "slow down" enough to intersect $2u$ again. $\endgroup$ – Arthur Aug 23 '13 at 7:20

In general, problems like this do not have closed-form solutions. Taking the logarithm of this equation; you get:

$$ x\log 8 = \log x + \log 6 $$

Generalize this to

$$ ax + b\log x + c = 0 $$

If this had a solution algebraic in $a, b, c: x = A(a, b, c)$, then

$$ \begin{align} aA + b\log A + c &= 0 \\ \log A &= -\frac{a}{b}A -\frac{c}{b} = B(a, b, c)\\ A(a, b, c) &= e^{B(a, b, c)} \end{align} $$

where both $A$ and $B$ are non-constant and algebraic in $a, b, c$. That is impossible.

I originally said that $\log x$ would be algebraic, and that isn't quite right.

This problem was devised for effect. We'll look backward here by seeing what happens if you substitute $y=3x$.

$$ \begin{align} 8^x &= 6x \\ 2^{3x} &=2 \cdot 3x\\ 2^y &= 2y \end{align} $$

Now, we consider the coincidence that $2^2 = 2\cdot2$ and $2^1 = 2\cdot 1$. So, $y=2$ and $y=1$ are solutions, and so $x=\frac23$ and $x=\frac13$ are solutions. But for general constants, you can't hope to guess at a solution. You will be in the realm of numerical approximation.

Not understanding how to solve this problem is not a sign of a lack of skill. The problem is too contrived.

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    $\begingroup$ Thanks @Eric for the nice and detailed explanation. +1 from me. $\endgroup$ – learner Aug 23 '13 at 7:11
  • $\begingroup$ Upvoted for that last sentence. $\endgroup$ – András Hummer Aug 23 '13 at 7:17
  • $\begingroup$ Nice solution Eric Jablow.... $\endgroup$ – juantheron Nov 9 '13 at 13:33

Hint: How many different roots can this equation have?

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    $\begingroup$ Infinitely many, if you allow complex roots. $\endgroup$ – Robert Israel Aug 23 '13 at 5:36
  • $\begingroup$ I know what you want to get at with your hint, but right now you are just rephrasing the question. Try to clarify, or else it's in trouble. $\endgroup$ – rschwieb Aug 23 '13 at 12:53
  • $\begingroup$ Right now the question is answered, so there is no need in hints. :-) $\endgroup$ – Alex Ravsky Aug 23 '13 at 18:46

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