Solving $\log_5 (2x+1)=\log_3 (3x-3)$. I am trying to resolve the equation $$\log_5 (2x+1) = \log_3 (3x-3)$$ and then of sketch the functions $y=\log_5 (2x+1)$ and $y=\log_3 (3x-3)$ get the solution $x=2$. There is an method that do not use the graphic method? Thanks for your suggestions. Any ideas will be appreciated.
 A: Let $$f(x) = \log_5 (2x+1)-\log_3(3x-3)$$ which is defined for $x>1$.
Since for $x>1$ we have $$(x+{1\over 2})\ln 5 > (x-1)\ln5 >(x-1)\ln3>0$$
we have also that $$f'(x) = {1\over (x+{1\over 2})\ln 5 } - {1\over (x-1)\ln3} <0$$
so given function has at most one zero ...
A: If you want a purely algebraic solution, although rather tedious, you can solve the equation as follows:
$$\log_5(2x+1) = \frac{\log_3(2x+1)}{\log_3 5} \tag{1}$$
$$\frac{\log_3(2x+1)}{\log_3 5} = \log_3(3x-3) \tag{2}$$
$$\left(3^{\log_3(2x+1)}\right)^{\frac{1}{\log_3 5}} = 3x-3 \tag{3}$$
$$(2x+1)^{\frac{1}{\log_3 5}} = 3x-3 \tag{4}$$
$$(2x+1)^{\log_5 3} = 3x-3 \tag{5}$$
$$\log_{2x+1} (3x-3) = \log_5 3 \tag{6}$$
$$\boxed{x = 2} \tag{7}$$
$(1)$: Change of bases: $\log_a b = \frac{\log_c b}{\log_c a}$
$(2)$: Rewriting the equation
$(3)$: Rewriting as an exponential: $c = \log_a b \iff a^c = b$
$(4)$: Simplifying
$(5)$: Inversion of base and argument: $\frac{1}{\log_a b} = \log_b a$
$(6)$: Rewriting as a logarithm: $a^b = c \iff \log_a b = c$
$(7)$: Solution to $2x+1 = 5$ and $3x-3 = 3$
But of course, you have to check for any extra solutions. Since $\log_a b^c = c\log_a b$ and $\log_{a^c} b = \frac{1}{c}\log_a b$, you have $$\log_{a^ c} b^c = \log_a b$$
$$\implies \log_5 3 = \log_{5^n} 3^n$$
$$\implies\log_{2x+1} (3x-3) = \log_{5^n} 3^n$$
$$\implies 2x+1 = 5^n; \quad 3x-3 = 3^n$$
$$3x-3 = 3^n \implies x-1 = 3^{n-1} \implies x = 3^{n-1}+1$$
Plugging this in the first equation, you have
$$2\left(3^{n-1}+1\right) = 5^n \implies 2\left(3^{n-1}\right)+3 = 5^n$$
It’s fairly obvious the only solution is $n = 1$ (since the RHS grows at a much faster rate), for which we already solved the equation. Hence, it’s the only solution.
