Question

If the terms $${\log }_{a}x,{\log }_{b}x,{\log }_{c}x$$ are in A.P where $$x\ne 1$$ , then show that $$c^2={\left(ac\right)}^{{\log }_{a}b}$$

Answer 1

Hint: The question is taken from the sequence and series from arithmetic progression and it is an arrangement of numbers in instruction in which the difference of any two consecutive numbers always remains constant value and in this given problem which we have to prove the given statement and if the three-term are in A.P. mean arithmetic progression we follow the following step consider a, b, c are in A.P then
$$b={\displaystyle \frac{a+c}{2}}$$
And since the given terms are logarithmic so we can use the logarithmic formula to obtain the statement.

Complete step-by-step solution:
Step 1:
First we write the given data or $${\log }_{a}x,{\log }_{b}x,{\log }_{c}x$$ in A.P. so it follow
a, b, c in A.P.
$$\Rightarrow$$$$b={\displaystyle \frac{a+c}{2}}$$
$${\log }_{a}x,{\log }_{b}x,{\log }_{c}x$$ in A.P. then
$$\Rightarrow$$$${\log }_{b}x={\displaystyle \frac{{\log }_{a}x+{\log }_{c}x}{2}}$$
Now use the logarithmic formula
$$\Rightarrow$$$${\log }_{p}q={\displaystyle \frac{{\log }_{w}q}{{\log }_{w}p}}$$
Apply the formula
$$\Rightarrow$$$${\log }_{b}x={\displaystyle \frac{{\log }_{a}x+{\log }_{c}x}{2}}$$
$$\Rightarrow$$$${\displaystyle \frac{{\log }_{c}x}{{\log }_{c}b}}={\displaystyle \frac{{\displaystyle \frac{{\log }_{c}x}{{\log }_{c}a}}+{\displaystyle \frac{{\log }_{c}x}{{\log }_{c}c}}}{2}}$$
$$\Rightarrow$$$${\displaystyle \frac{{\log }_{c}x}{{\log }_{c}b}}={\displaystyle \frac{{\log }_{c}x\left[{\displaystyle \frac{1}{{\log }_{c}a}}+{\displaystyle \frac{1}{{\log }_{c}c}}\right]}{2}}$$
$$\Rightarrow$$$${\displaystyle \frac{1}{{\log }_{c}b}}={\displaystyle \frac{\left[{\displaystyle \frac{1}{{\log }_{c}a}}+{\displaystyle \frac{1}{{\log }_{c}c}}\right]}{2}}$$
$$\Rightarrow$$$${\displaystyle \frac{1}{{\log }_{c}b}}={\displaystyle \frac{1}{2}}\left[{\displaystyle \frac{{\log }_{c}c+{\log }_{c}a}{{\log }_{c}a{\log }_{c}c}}\right]$$
Apply the formula
$$\Rightarrow$$$${\log }_{q}p+{\log }_{q}q={\log }_{q}pq,{\log }_{q}q=1$$
$$\Rightarrow$$$${\displaystyle \frac{1}{{\log }_{c}b}}={\displaystyle \frac{1}{2}}\left[{\displaystyle \frac{{\log }_{c}ac}{{\log }_{c}a}}\right]$$
$$\Rightarrow$$$${\displaystyle \frac{{\log }_{c}a}{{\log }_{c}b}}=\left[{\displaystyle \frac{{\log }_{c}ac}{2}}\right]$$
Now again apply the formulas
$$\Rightarrow$$$${\log }_{p}q={\displaystyle \frac{{\log }_{w}q}{{\log }_{w}p}},{\log }_{p}q={\displaystyle \frac{1}{{\log }_{q}p}},{\log }_p{q}^n=n{\log }_{p}q$$
$$\Rightarrow$$$${\log }_{b}a=\left[{\displaystyle \frac{{\log }_{c}ac}{2}}\right]$$
$$\Rightarrow$$$${\displaystyle \frac{1}{{\log }_{a}b}}=\left[{\displaystyle \frac{1}{2{\log }_{ac}c}}\right]$$
$$\Rightarrow$$$${\log }_{a}b=2{\log }_{ac}c$$
$$\Rightarrow$$$${\log }_{a}b={\log }_{ac}{c}^2$$
Now we know that
$$\Rightarrow$$$${\log }_{q}p=k\Rightarrow p=q^k$$ Then
$$\Rightarrow$$$${\log }_{ac}{c}^2={\log }_{a}b$$
$$\Rightarrow$$$$c^2={\left(ac\right)}^{{\log }_{a}b}$$
We proved the given statement

Note: The best part about basic logarithmic is that it can be fit anywhere and the given problem is complex but the basic formula of the logarithmic can be solved easily hence we proved the given statement in the above solution that is provided step by step. Since the given question from the sequence and series of arithmetic progression and we only apply the single concept that is considered a, b, c are in A.P then
$$b={\displaystyle \frac{a+c}{2}}$$
And after we applied the concept that it becomes a logarithmic-based problem and we have solved the problem easily and proved the given statement as mentioned above.