An algebra problem by akwesi barnes

$\ Ln(x^2) = Ln(16^x) \implies\ 2Ln|x| = xLn16$ Note that the modulus sign in the first logarithm is significant as both $\ x^2$ and $\ 16^x$ are both positive when $\ x$ is negative. Sketching both graphs on one axis shows that $\ x = -0.5$ is a solution, but it may be unclear whether there are any positive solutions (unless you used Wolfram Alpha/autograph/geogebra etc..). Now let $\ f(x) = 2Ln(x) \ and \ g(x) = x Ln(16)$ . Now the gradient of g(x) is $\ g'(x) = Ln(16)$ and $\ f'(x) = \frac{2}{x}$ . Comparing gradients: $\ f'(x) > g'(x) \Rightarrow\ x Ln(16) > \frac{2}{x} \Rightarrow\ x > \sqrt{\frac{2}{Ln(16)}} \approx 0.85$ Plugging is x = 0.85 into both functions shows that g(0.85) > f(0.85) so the gradient of both functions tells us that $\ g(x) > f(x) \forall x> 0$ . Hence, $\ x = -0.5$ is a unique (real) solution.

logx^2=log16^x

2logx=xlog16

So... x^2 = (4^2)^x x^2 = (4^x)^2 X = 4^x X^1/x = 4 the only possible solution is -1/2 or - 0.5