Express the logarithm of number 63 to the base 140 in terms of a,b and c

logarithms

I'm solving logs question. And got stuck in this question.

Given that $\log_2 3 = a$, $\log_3 5 = b$ and $\log_7 2 = c$, express the logarithm of number 63 to the base 140 in terms of a,b and c

I've tried to sole it. But wasn't able to move after it.
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Please explain, how I can solve it and please solve it according to class 11th student.

Best Answer

A way to make our lives somewhat more difficult: $$\log_23=a\Rightarrow2^a=3\\\log_35=b\Rightarrow3^b=5\\\log_72=c\Rightarrow7^c=2\Rightarrow 7=2^{\frac1c}$$

Let $x=\log_{140}63\Rightarrow140^x=63$

Notice that $140=20*7=2*2*5*7$ and $63=7*9=3*3*7$

So we write everything dependent on $2, 5$ and $c$. $$(2*2*5*7)^x=2^a*2^a*2^{\frac1c}\Rightarrow\\(2*2*5*2^{\frac1c})^x=2^a*2^a*2^{\frac1c}\Rightarrow\\x\log_2(2^{\frac1c+2})+\log_25=2a+\frac1c\Rightarrow\\x=c\frac{[2a-\log_25]+1}{2c+1}$$

Technically we get rid of $b$, though we have the $\log_25$ in the expression..

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