Solve logarithmic equation $x-a=x^{\log_{10}2}$

Question

I have the following equation:

$$x-8=x^{\log_{10}2}$$

I saw that $x=10$ but I don't know how to solve it. So?

Answer 1

hint: we can write $$\ln(x-8)=\frac{\ln(2)}{\ln(10)}\ln(x)$$

Answer 2

Just for your curiosity, assuming $0<b<1$, let us consider the function and its derivatives $$f(x)=x-x^{b}-a$$ $$f'(x)=1-b\, x^{b-1}$$ $$f''(x)=(1-b) \,b \,x^{b-2}$$ The first derivative cancels at $$x_*=\left(\frac{1}{b}\right)^{\frac{1}{b-1}}>0$$ and the second derivative test shows that this corresponds to a minimum of the function.

As Kibble answered, for $a\neq 8$, there is no analytical solution to this equation and numerical method should be used. Newton method is probably the simplest to use. Starting from a reasonable guess $x_0$, the method will update it according to $$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$$ For the case you consider ($b=\log_{10}(2)$), you undertand that, as soon as $x>1$, the function is essentially defined by $x$, that is to say that $x_0=a$ is probably a good starting point. Let us try with $a=123$; applying Newton method, the successive iterates will then be $$x_1=127.302023786721$$ $$x_2=127.301480972598$$ $$x_3=127.301480972590$$ which is the solution for fifteen significant figures.

Hoping that this could be of some interest.