Need help solving this log equation $\log_{5}(x-4) = \log_{10}(x+5)$

Question

I need help solving for $x$:

$$\log_{5}(x-4) = \log_{10}(x+5)$$

Answer 1

Just to help close this question, let me demonstrate the difficulty in solving this exactly. "$\lg$" denotes the base $10$ logarithm.

By change of base, we can transform the LHS. The equation becomes:

$\frac{\lg(x-4)}{\lg 5} = \lg (x+5)$

$\lg(x-4) = \lg(x+5)^{\lg 5}$

Take inverse logs of both sides:

$x-4 = (x+5)^{\lg 5}$

And that's where you have to throw in the towel. That is a transcendental equation, no way to simplify that algebraically with elementary functions. Expressing $\lg 5 = 1 - \lg 2$ won't help.

The only time when you can solve your form of equation exactly is when you have the same expression in $x$ on either side. So, for instance, if you had $x+5$ on both sides, you could let $y=x+5$ and write:

$\lg y = \lg 5 \cdot \lg y$

Giving us the immediate solution by inspection: $\lg y = 0 \implies y = 1 \implies x = - 4$. No other solution exists.

Answer 2

Try to convert $\log_5 K = \log_{2\cdot 5=10} M$ how does that work.

$\log_5 A = k$ means $5^k = A$ so $10^k = \frac {A}{2^k}$ so $\log_{10}\frac {A}{2^{\log_5 A}} = \log_5 A$

So $\log_{10}\frac {x-4}{2^{x-4}} = \log_{10} (x+5)$

So $\frac {x-4}{2^{x-4}} = x+5$.

$x-4 = (x+5)2^x\frac 1{16}$

$16\frac {x-4}{x+5} = 2^x$

$1- \frac 9{x+5} = 2^{x-4}$ and ... I quit.

You can google how to solve these with lambert functions but.... I quit.

Hmm... actually $1-2^{x-4} = \frac 9{x+5}$ gives us a good test to guess by trial and error..... but I still quit.

Answer 3

Using natural logarithms and simplifying, you are looking for the zero's of function $$f(x)=\log (10) \log (x-4)-\log (5) \log (x+5)$$ Its first derivative $$f'(x)=\frac{\log (10)}{x-4}-\frac{\log (5)}{x+5}$$ cancels at $$x_*=\frac{4 \log (5)+5 \log (10)}{\log (5)-\log (10)}\sim -25.8974$$ which is lower than $4$. So, the first derivative is always positive and we have an increasing function.

Now, if you plot or use inspection, you will notice that the single root is close to $x=11$. Use it as the $x_0$ for Newton method which will generate the following iterates $$\left( \begin{array}{cc} n & x_n \\ 0 & 11.00000000 \\ 1 & 10.91979756 \\ 2 & 10.92036742 \\ 3 & 10.92036745 \end{array} \right)$$