Equation with $x$ in a logarithm and exponent

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.

• Is $\log$ base 10 or natural? – Andrew Li Jun 6 '18 at 21:53
• 10, sorry I thought log with no base defaults to 10 always – Insanit Jun 6 '18 at 21:55
• It's common in higher level math for $\log$ to mean the natural logarithm. So typically one writes $\log_{10}$ for the base-10 logarithm to be unambiguous. – arkeet Jun 6 '18 at 21:57
• OK, thanks I will remember that. But what's the point of that convention if there is ln(x) to mean natural logarithm? – Insanit Jun 6 '18 at 22:04
• Ooog. Don't get us started. Mathematicians are a prickly bunch. That's the point of the convention. ... Okay, mathematicians don't believe $10$ has any significance at all and the only log that does is the natural log and because natural log is the norm, not the exception, to have a terminology $\ln$ is perverse and unnecessary.... Look, just ... let it go.... – fleablood Jun 7 '18 at 6:12

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.

• What does IVT mean? Also I understand that there exists a solution for x, but what exactly is the "numerical method" for finding it? – Insanit Jun 6 '18 at 22:01
• – gimusi Jun 6 '18 at 22:03
• ok I get that there is a solution but could you please show step by step for how to algebraically solve this equation? – Insanit Jun 6 '18 at 22:06
• @Insanit We can't find explicit derivation bt elementary functions, we need numerical methods to obtain the solution. – gimusi Jun 6 '18 at 22:12
• I'm not sure what that means. Could you please rephrase that in layman's terms? Remember that I am in Grade 11, I'm sorry if it's frustrating to explain to me :( – Insanit Jun 6 '18 at 22:14

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 $16.000000000178704123062$