How to solve this equation using logs

Question

How do solve this equation for x using logarithms?

$$4^x = 6^x-3$$

If it is not possible using logarithms, please provide another way. Thank you in advance

Answer 1

The solution is roughly $x=1.1602$, but this is a transcendental equation with no obvious tricks to solve it exactly. If you had $$4^x = 3\cdot 6^x$$ then you could solve it by taking the log of both sides and solving for $\log x$.

Answer 2

In the most general case, equations such as $a^x=b^x+c$ do not show explicit solutions and numerical methods (such as Newton) must be used. Considering that we look for the zero of function $$f(x)=a^x-b^x-c$$ starting from a reasonable guess $x_0$, Newton method will update the guess according to $$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$$ Let us consider the case of your equation $$f(x)= 6^x-4^x-3$$ $$f'(x)=6^x \log (6)-4^x \log (4)$$ By inspection, $f(1)=-1$, $f(2)=+17$ show that the solution is quite close to $1$; so, let us use $x_0=1$ and apply the method. This will generate the following iterates $$x_1=1.192109$$ $$x_2=1.161257$$ $$x_3=1.160180$$ $$x_4=1.160179$$ which is the solution for seven significant figures.

However, we can do it faster considering instead $$g(x)=\log(6^x)-\log(4^x+3)=x\log(6)-\log(4^x+3)$$ which is much more linear. Doing the same would lead to the following iterates $$x_1=1.154214$$ $$x_2=1.160170$$ $$x_3=1.160179$$