I am stuck on the following problem which one of my friends gave me:
Solve : $\,8^x=6x$.
MY ATTEMPTS:
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.
$\endgroup$ 24 Answers
$\begingroup$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 $x=2/3$.
$\endgroup$ 1 $\begingroup$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.
$\endgroup$ 3 $\begingroup$Hint: How many different roots can this equation have?
$\endgroup$ 3 $\begingroup$$8^x=6x\implies e^{x\log 8}=6x\implies -xe^{-x\log8}\log8=-\dfrac{\log8}6$
Recall the Lambert W-function is defined as the inverse of $ze^z$ in $\Bbb C$, so if we let $t=-x\log8$, then $te^t=-\dfrac{\log8}6\implies t=W(-\dfrac{\log8}6)$
Hence, $x=-\dfrac{W(\dfrac{\log8}6)}{\log8}$.
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