# How to solve : $\,8^x=6x$

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.

• Well, the Lambert-W function might help here: en.wikipedia.org/wiki/Lambert_W_function#Example_1 ? Aug 23, 2013 at 4:29
• I tried to solve this without using derivatives only to get frustrated. Oh well. Aug 23, 2013 at 4:31

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 $x=2/3$.

• @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. Aug 23, 2013 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.

• Thanks @Eric for the nice and detailed explanation. +1 from me. Aug 23, 2013 at 7:11
• Upvoted for that last sentence. Aug 23, 2013 at 7:17
• Nice solution Eric Jablow.... Nov 9, 2013 at 13:33

Hint: How many different roots can this equation have?

• Infinitely many, if you allow complex roots. Aug 23, 2013 at 5:36
• 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. Aug 23, 2013 at 12:53
• Right now the question is answered, so there is no need in hints. :-) Aug 23, 2013 at 18:46

$$8^x=6x\implies e^{x\log 8}=6x\implies -xe^{-x\log8}\log8=-\dfrac{\log8}6$$

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, $$x=-\dfrac{W(\dfrac{\log8}6)}{\log8}$$.