Algebraic solution to

Question

The equation $$16^{x} = x^{2}$$ has an obvious solution of $x = -\dfrac{1}{2}$. However, I can't find algebraic solution to demonstrate this answer using logarithm and index laws. Any help is much appreciated.

Answer 1

We have

$16^x = x^2$

$x\ln16 = 2\ln |x|$

$\frac{\ln|x|}{x} = \ln 4$

For $x>0$, we have that $\frac{\ln x}{x}$ has a maximum of $\frac1e < \ln 4$ so the equation has no solutions for $x > 0$.

For $x<0$, we have that $\frac{\ln t}{t} = \ln\frac14$ where $t>0$. As $\ln\frac14 < 0$, this has only one solution in $t$ which is $t=\frac12$.

The only solution of the given equation is $x=-\frac12$.

Answer 2

One way of illustrating the laws you mention is as follows.

From $16^x=x^2$ we have either $4^x=x$ or $4^x=-x$.

From the graph of $y=4^x$ the first possibility can be seen to have no solutions and so $4^x=-x$.

Let $x=-2^{n-1}$ and take logs to the base 2.

Then $-2^{n-1}\log 4=(n-1)\log 2$ and so $$n+2^{n+1}=1.$$

We now have two monotonically increasing functions whose sum is 1 only when $n=0$ i.e. $x=-\frac{1}{2}$.