I came across this logarithm equation and can't seem to figure out how to solve it.

$$lg(2x-24)=2+ \frac 13 lg8 - \frac 14 x lg16$$

I only managed to simplify all the way till $$100=(x-12)(2^x)$$ (based on assumption of base 10)

Would gladly appreciate if anyone could suggest a trick to solve this.

Edit: Could give some benefit of doubt to whether the log is base 2 or 10. Any solutions are welcomed.

  • $\begingroup$ Just to confirm, is this supposed to be log base 10? $\endgroup$
    – 2012ssohn
    Aug 17, 2017 at 5:10
  • $\begingroup$ @2012ssohn I presume $\lg$ is base two logarithm, but then I cannot see where $100$ comes from. $\endgroup$ Aug 17, 2017 at 5:15
  • $\begingroup$ There is no obvious integer solution to your last equation, so I'm tempted to say it's unsolvable (equations usually are when you have $x$ both as an exponent and as a "non-exponent", for lack of a better word). $\endgroup$
    – Arthur
    Aug 17, 2017 at 5:16
  • $\begingroup$ @2012ssohn The question didn't state but I would believe it to be base 10. I used base 10 to obtain 100. $\endgroup$ Aug 17, 2017 at 5:17
  • $\begingroup$ @Arthur What about if any real solutions can be accepted? $\endgroup$ Aug 17, 2017 at 5:20

3 Answers 3


Assuming your logs are base $2$ we have $$\lg(2x-24)=2+ \frac 13 \lg8 - \frac 14 x \lg16\\ \lg(x-12)+1=2+1-x\\ lg(x-12)=2-x\\x-12=2^{2-x}\\ 2^x(x-12)=4$$ Clearly $x$ has to be just barely greater than $12$, so let $x=12+y$ Then we have $$y=\frac 1{1024\cdot 2^y}$$ This needs a numeric solution and is in a good form as the right side will change slowly with $y$. Let $y_0=0$ and iterate. After two iterations we have converged to $0.000975902$. Alpha will give you a solution of $y=\frac {W\left(\frac {\log 2}{1024}\right)}{\log 2}$in terms of the Lambert W function where these logs are natural logs.

Added: for base $10$ logs we can do the same. Again $x$ has to be a little greater than $12$ so write $x=12+y$ The equation becomes $$y=\frac {25}{1024\cdot 2^y}$$ which converges to $y\approx 0.0240$

  • 1
    $\begingroup$ Is it possible to solve without such numerical methods? Assuming knowledge of math up to high school level. $\endgroup$ Aug 17, 2017 at 5:27
  • 1
    $\begingroup$ Mixes of exponentials and polynomials are usually not soluable without the W function or numerics. $\endgroup$ Aug 17, 2017 at 5:30

This is indeed a very good introduction to Lambert function.

Sooner or later, you will learn that any equation which can write or rewrite $$A+Bx+C\log(D+Ex)=0$$ has solution(s) which espress(es) in terms of this function.

In the case of natural logarithms, using the steps shown in Ross Millikan's answer, you would end with $$x=12+\frac{1}{\log_e (2)}W\left(\frac{e^2 \log_e (2)}{4096}\right)$$

Assuming logarithms in base $2$ as the numbers suggest, then, as Ross Millikan answered, $$x=12+\frac{1}{\log_e (2)}W\left(\frac{\log_e (2)}{1024}\right)$$

Assuming logarithms in base $10$, $$x=12+\frac{1}{\log_e (2)}W\left(\frac{25 \log_e (2)}{1024}\right)$$

Now, since the argument is quite small, you can approximate the value of $W(t)$ using the expansion $$W(t)=t-t^2+\frac{3 }{2}t^3-\frac{8 }{3}t^4+O\left(t^5\right)$$ or, better, using Padé approximants such as $$W(t)=\frac{t }{1+t }$$ $$W(t)=\frac{t+\frac{4}{3} t^2}{1+\frac{7 }{3}t+\frac{5 }{6}t^2 }$$


Once you arrive at $100$ = ($x−12$)($2^x$) ,

$x-12$ = $100$*$(2^{-x})$ ,

x = 100*(2^(-x)) + 12 ,

On my calculator, to get an approximation, i just started with some random number ($1$ in my case), and entered $100$*($2^{-ans}$) + $12$ , where ans is the previous value. By repeatedly evaluating this, it converges to to the value of $x$ = $12.0240111$. In theory, this should work from any initial value.


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