Logarithm abc

The given relations can be rewritten in terms of natural logarithms using the following identity:

$\frac{\ln{3}}{\ln{2}} = a \dots (1)$ $\frac{\ln{5}}{\ln{3}} = b \dots (2)$ $\frac{\ln{15}}{\ln{7}} = c \dots (3)$

Multiplying (1) and (2): $\frac{\ln{5}}{\ln{2}} = ab \dots (4)$

$c = \frac{\ln{15}}{\ln{7}} = \frac{\ln{5} + \ln{3}}{\ln{7}}$ $c = \frac{\frac{\ln{5}}{\ln{2}} + \frac{\ln{3}}{\ln{2}}}{\frac{\ln{7}}{\ln{2}}}$ $\implies c = \frac{ab + a}{\frac{\ln{7}}{\ln{2}}}$ $\implies \frac{\ln{7}}{\ln{2}} = \frac{ab + a}{c} \dots (5)$

Now:

$\frac{\ln{45}}{\ln{14}} = \frac{\ln{5} + 2\ln{3}}{\ln{7} + \ln{2}}$ $\frac{\ln{45}}{\ln{14}} = \frac{\frac{\ln{5}}{\ln{2}} + 2\frac{\ln{3}}{\ln{2}}}{\frac{\ln{7}}{\ln{2}}+ 1}$

Using relations (1), (4) and (5): $\frac{\ln{45}}{\ln{14}} = \frac{ab + 2a}{\frac{ab + a}{c}+ 1}$ $\boxed{\frac{\ln{45}}{\ln{14}} = \frac{(2+b)ac}{a+ab+c}}$

We'll use a couple of logarithmic identities: $\log_xx=1$ $\log_x(yz)=\log_xy+\log_xz$ $\log_zy=\frac{\log_xy}{\log_xz}$ $\log_xy=\frac{1}{\log_yx}$ The last one follows from the first and the third identity. Now let's rework what we're given: $\log_{14}45=\log_{14}15+\log_{14}3$ $=\frac{1}{\log_{15}14}+\frac{1}{\log_314}$ $=\frac{1}{\log_{15}7+\log_{15}2}+\frac{1}{\log_32+\log_37}$ $=\frac{1}{\frac{1}{c}+\frac{1}{\log_215}}+\frac{1}{\frac{1}{a}+\frac{\log_{15}7}{\log_{15}3}}$ $=\frac{1}{\frac{1}{c}+\frac{1}{\log_23+\log_25}}+\frac{1}{\frac{1}{a}+\frac{\log_315}{\log_715}}$ $=\frac{1}{\frac{1}{c}+\frac{1}{a+\frac{\log_35}{\log_32}}}+\frac{1}{\frac{1}{a}+\frac{\log_33+\log_35}{c}}$ $=\frac{1}{\frac{1}{c}+\frac{1}{a+ab}}+\frac{1}{\frac{1}{a}+\frac{1+b}{c}}$ After careful simplifying we reach to the answer: $\log_{14}45=\frac{(b+2)ac}{a+ab+c}$

$\large \begin{aligned} \log_{14} 45 & = \frac {\log (3^2\times 5)}{\log (2 \times 7)} \\ & =\frac {2 \log 3 + \log 5}{\log 2 + \log 7} & \small \blue{\text{Divide up and down by }\log 2} \\ & = \frac {2 \blue{\frac {\log 3}{\log 2}} + \frac {\log 5}{\log 2}}{\frac {\log 2}{\log 2} + \frac {\log 7}{\log 2}} & \small \blue{\text{Note that }a = \log_2 3 = \frac {\log 3}{\log 2}} \\ & = \frac {2\blue a+ \frac \red{\log 5}\blue{\log 2} \times \frac \blue{\log 3}\red{\log 3}}{1 + \frac \blue{\log 7}{\log 2} \times \frac {\log 15}\blue{\log 15}} & \small \red{\text{Similarly }b = \frac {\log 5}{\log 3}} \\ & = \frac {2\blue a+ \frac \red{\log 5}\blue{\log 2} \times \frac \blue{\log 3}\red{\log 3}}{1 + \frac \blue{\log 7}{\log 2} \times \frac {\log 15}\blue{\log 15}} & \small \blue{\text{and }c = \frac {\log 15}{\log 7}} \\ & = \frac {2\blue a + \blue a \red b}{1+ \frac {\log 3 + \log 5}{\blue c \log 2}} \\ & = \frac {2a + ab}{1+ \frac ac + \frac {\log 5}{c \log 2} \times \frac {\log 3}{\log 3}} \\ & = \frac {2a + ab}{1+ \frac ac + \frac {ab}c} \\ & = \boxed{\frac {(2+b)ac}{a+ab+c}} \end{aligned}$