Equation with $x$ in a logarithm and exponent

Question

Solve for $x$: $3 \log_{10}(x-15) = \left(\frac{1}{4}\right)^x$

I seem to get stuck when I get to logarithm of a logarithm or power of a power, graphing it and doing some guess and check on the calculator shows that $x$ should be just a bit above 16, but I would like to know how to figure it out algebraically if possible. I'm in Grade 11 so I probably won't understand anything too complicated.

Answer 1

Note that

• $f(x)=3\log_{10} (x-15)$, defined for $x>15$, is strictly increasing

• $g(x)=\frac1{4^x}$ is strictly decreasing

and

• $f(16)=0<g(16)$

• $f(25)=3>g(25)$

then by IVT a solution exists for $x\in(16,25)$ which can be found by numerical methods.

Answer 2

Consider that you look for the zero of $$f(x)=3 \log_{10}(x-15) - \left(\frac{1}{4}\right)^x=\frac{3 }{\log (10)}\log (x-15)-4^{-x}$$ $$f'x)=\frac{3}{(x-15) \log (10)}+4^{-x} \log (4)$$ Since you notice that $x$ should be just a bit above $16$, perform one single iteration of Newton method writing $$0=f(16)+f'(16)(x-16)\implies x=16-\frac{f(16)}{f'(16)}$$ This should give $$x=16+\frac{1}{\log (4)+\frac{12884901888}{\log (10)}}\approx 16.000000000178704123046$$ while the "exact" solution would be $16.000000000178704123062$