1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Is $\log$ base 10 or natural? $\endgroup$ – Andrew Li Jun 6 '18 at 21:53
  • $\begingroup$ 10, sorry I thought log with no base defaults to 10 always $\endgroup$ – Insanit Jun 6 '18 at 21:55
  • $\begingroup$ 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. $\endgroup$ – arkeet Jun 6 '18 at 21:57
  • $\begingroup$ OK, thanks I will remember that. But what's the point of that convention if there is ln(x) to mean natural logarithm? $\endgroup$ – Insanit Jun 6 '18 at 22:04
  • $\begingroup$ 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.... $\endgroup$ – fleablood Jun 7 '18 at 6:12
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ What does IVT mean? Also I understand that there exists a solution for x, but what exactly is the "numerical method" for finding it? $\endgroup$ – Insanit Jun 6 '18 at 22:01
  • $\begingroup$ @Insanit en.wikipedia.org/wiki/Intermediate_value_theorem $\endgroup$ – gimusi Jun 6 '18 at 22:03
  • $\begingroup$ ok I get that there is a solution but could you please show step by step for how to algebraically solve this equation? $\endgroup$ – Insanit Jun 6 '18 at 22:06
  • $\begingroup$ @Insanit We can't find explicit derivation bt elementary functions, we need numerical methods to obtain the solution. $\endgroup$ – gimusi Jun 6 '18 at 22:12
  • $\begingroup$ 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 :( $\endgroup$ – Insanit Jun 6 '18 at 22:14
2
$\begingroup$

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 $16.000000000178704123062$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.