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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.
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 $y\approx 0.0240$
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.
