The b is supposed to be lowercase in the log functions but I do not know how to do that yet in this syntax.
$1)$ Find $x - y$
where $x = 2^{\log_b(3)}$ and $y = 3^{\log_b(2)}$
$2^{\log_b(3)} = \log_b(3)\ln(2)$
=> $\log_b(3) = \dfrac{1}{\ln(2)}$
=> $b^{\frac{1}{\ln(2)}} = 3$
This is where I got stuck. I need to find b but am confused on how to find it in this situation. Once I can find b in x I'm able to solve it for y and then just compute x - y since I know the base.
$2)$
if $\dfrac{\log_b(a)}{\log_c(a)} = \dfrac{19}{99}$ and $\dfrac{b}{c} = c^k$, Find $K$.
$\dfrac{\log_b(a)}{\log_c(a)} = \log_b(a) - \log_c(a)$
I'm not quite sure about this question because I have to still review the logarithmn identities but any ideas would be appreciated for this.
If I'm not mistaken these solutions are correct
For Problem $1$. $$x=2^{\log_{b}3}\implies \log x=\log_{b}{3}\,\log2=\log2\frac{\log 3}{\log b}=\frac{\log2\log 3}{\log b}$$ $$y=3^{\log_{b}2}\implies \log y=\log_{b}2\log3=\log3\frac{\log2}{\log b}=\frac{\log2\log 3}{\log b}$$
$$\implies \log x=\log y\implies x=y$$
For Problem $2$.
$$\dfrac{b}{c} = c^k\implies b=c^{k+1}$$ $$\dfrac{\log_b(a)}{\log_c(a)} = \dfrac{19}{99}=\frac{\log a}{\log b}\cdot\frac{\log c}{\log a}=\frac{\log c}{\log b}=\frac{\log c}{\log c^{k+1}}=\frac{\log c}{(k+1)\log c}=\frac{1}{k+1}=\frac{19}{99}$$