Question

If \[{\log _4}5 = a\]and \[{\log _5}6 = b\], then find the value of \[{\log _3}2\]

Answer 1

Hint: We are going to use one of the logarithmic rules named the change of base rule. Applying this rule on the given problem and simplifying will directly lead us to the solution.

Formulae used:
1) Change of Base Rule: \[{\log _a}b = \dfrac{{\log b}}{{\log a}}\]
2) The Product Rule: \[\log ab = \log a + \log b\]
3) \[\log {a^b} = b\log a\]

Complete step-by-step answer:
It is given that \[{\log _4}5 = a\]and \[{\log _5}6 = b\]
First we are going to apply the logarithmic change of base rule on \[{\log _4}5\]
\[{\log _4}5 = \dfrac{{\log 5}}{{\log 4}}\]
\[\dfrac{{\log 5}}{{\log 4}} = a\](By given)
Now we are going to cross multiply.
\[\log 5 = a\log 4\]
Now we know that 4 can be rewritten as \[{2^2}\].So we can replace 4 by \[{2^2}\]
So we have, \[\log 5 = a\log {2^2}\]
Now we are going to use the logarithmic formula \[\log {a^b} = b\log a\]on the above equality.
\[ \Rightarrow \log 5 = 2a\log 2...........(i)\]
Now we are going to apply the logarithmic change of base rule on \[{\log _5}6\]
          \[{\log _5}6 = \dfrac{{\log 6}}{{\log 5}}\]
\[ \Rightarrow \dfrac{{\log 6}}{{\log 5}} = b\](By given)
Now we are going to replace 6 by $2 \times 3$
\[ \Rightarrow \dfrac{{\log (2 \times 3)}}{{\log 5}} = b\]
Now we are going to apply the logarithmic product rule on \[\log (2 \times 3)\].
\[\log (2 \times 3) = \log 2 + \log 3\].
So we have, \[\dfrac{{\log 2 + \log 3}}{{\log 5}} = b\]
Now we are going to substitute (i) on the above equation. We are going to replace log5 by 2a log2 in the above equation.
\[ \Rightarrow \dfrac{{\log 2 + \log 3}}{{2a\log 2}} = b\]
Now we are going to cross multiply.
\[ \Rightarrow \dfrac{{\log 2 + \log 3}}{{\log 2}} = 2ab\]
Now we are going to split the denominator term.
\[ \Rightarrow \dfrac{{\log 2}}{{\log 2}} + \dfrac{{\log 3}}{{\log 2}} = 2ab\]
Now we cancel the term log 2 on both numerator and denominator.
\[ \Rightarrow 1 + \dfrac{{\log 3}}{{\log 2}} = 2ab\]
Now we are going to subtract 1 on both sides. Then we get
\[ \Rightarrow \dfrac{{\log 3}}{{\log 2}} = 2ab - 1\]
Now we are going to apply the change of base rule on \[\dfrac{{\log 3}}{{\log 2}}\].
So we have, \[\dfrac{{\log 3}}{{\log 2}} = {\log _2}3\]
\[ \Rightarrow {\log _2}3 = 2ab - 1\]
Now we know that \[{\log _3}2 = \dfrac{{\log 2}}{{\log 3}}\] (By change of base rule)
\[{\log _3}2 = \dfrac{{\log 2}}{{\log 3}} = \dfrac{1}{{\dfrac{{\log 3}}{{\log 2}}}}\]
We already know that the value of \[\dfrac{{\log 3}}{{\log 2}}\] is \[2ab - 1\]. So we are going to substitute that value on the above equation.
\[ \Rightarrow {\log _3}2 = \dfrac{1}{{2ab - 1}}\]
Hence the value of \[{\log _3}2\] is \[\dfrac{1}{{2ab - 1}}\]

Note:
Like exponents, there are logarithmic properties that allow us to simplify and modify the given logarithmic expressions. So one must know these properties to get the desired solution.