If \[{\log _8}m = 3.5\] and ${\log _2}n = 7$, then find the value of m in terms of n.

Answer Verified Verified
Hint: We will use the logarithmic identity ${\log _a}x = b \Rightarrow x = {a^b}$ to solve the given problem for finding the value of m in terms of n. Firstly we will calculate the values of m and n from the given logarithmic equations using the logarithmic identity mentioned above and then we will solve them for generalising the value of m in terms of n.

Complete step-by-step answer:
We are given two logarithmic equations: \[{\log _8}m = 3.5\] and ${\log _2}n = 7$.
We have a logarithmic identity: ${\log _a}x = b \Rightarrow x = {a^b}$. Using this identity in the given equations, we can calculate the values of m and n.
Using the identity in the equation: \[{\log _8}m = 3.5\], we get
\[ \Rightarrow {\log _8}m = 3.5 \Rightarrow m = {8^{3.5}}\]
Now, using the logarithmic identity in the equation: ${\log _2}n = 7$, we get
$ \Rightarrow {\log _2}n = 7 \Rightarrow n = {2^7}$
Now, when we have the values of m and n as: m = ${8^{3.5}}$and n =${2^7}$, we can write m = ${8^{3.5}}$ as $m = {8^{\dfrac{{35}}{{10}}}}$.
Now, we have $m = {8^{\dfrac{{35}}{{10}}}}$. It can be re – written as: $m = {\left( {{2^3}} \right)^{\dfrac{7}{2}}}$
Now, we can exchange the powers of 2 by using the algebraic property ${({x^a})^b} = {({x^b})^a}$.
$ \Rightarrow m = {\left( {{2^7}} \right)^{\dfrac{3}{2}}}$
Now, we know that the value of n = ${2^7}$, therefore, putting the value of n in the equation of m, we get
$ \Rightarrow m = {\left( n \right)^{\dfrac{3}{2}}}$
Or, we can write this as $m = n\sqrt n $.

Therefore, the value of m in terms of n is: $m = n\sqrt n $.

Note: In such problems, you may get confused about the base change rule of logarithmic functions and also after calculation of value of m and n, you may get caught up in how to proceed further. Also, the identity used should be known to you may it be logarithmic or algebraic identity. This will help you to understand the procedure to solve questions efficiently and much quickly.