Question

If \[{\log _k}x \cdot {\log _5}k = {\log _x}5,k \ne 1,k > 0\], what is the value of \[x\] ?
A. \[k\]
B. 0
C. 5
D. None of these

Answer 1

Hint: First, we use logarithm formula of \[{\log _b}a = \dfrac{{\log a}}{{\log b}}\] and simplifying the given equation. After that, we use cross multiplication to reach a more simplified form of logarithm. We use the antilog formula to reach the required solution.

Formula used:
Logarithm formula, \[{\log _b}a = \dfrac{{\log a}}{{\log b}}\]
Antilog formula, if \[\log a = \log b\] , where \[a,b \in \mathbb{N}\] then \[a = b\] .
Square root, \[\sqrt {{a^2}} = a\] , where \[a\] is an integer.
Complete step by step Solution:
Here we are given the equation
\[{\log _k}x \cdot {\log _5}k = {\log _x}5,k \ne 1,k > 0\]……………(1)
As we know about the logarithmic property
\[{\log _b}a = \dfrac{{\log a}}{{\log b}}\]
We will apply it in the given equation (1), which gives us
\[ \Rightarrow \left( {\dfrac{{\log x}}{{\log k}}} \right)\left( {\dfrac{{\log k}}{{\log 5}}} \right) = \dfrac{{\log 5}}{{\log x}}\]
Multiplying left side of the above equation and we get
\[ \Rightarrow \left( {\dfrac{{\log x}}{{\log k}} \times \dfrac{{\log k}}{{\log 5}}} \right) = \dfrac{{\log 5}}{{\log x}}\]
Simplifying and we get
\[ \Rightarrow \dfrac{{\log x}}{{\log 5}} = \dfrac{{\log 5}}{{\log x}}\]
Cross multiplying the above equation, we get
\[ \Rightarrow {(\log x)^2} = {(\log 5)^2}\]
Applying square root and we get
\[ \Rightarrow \log x = \log 5\]
Applying antilog formula and we get
\[ \Rightarrow x = 5\]
Hence, the correct option is option C.

Note: Students got confused with the logarithm formula and square root of an expression. If we use \[{\log _b}a = \dfrac{{\log b}}{{\log a}}\] as logarithm formula instead of \[{\log _b}a = \dfrac{{\log a}}{{\log b}}\] , then we got wrong answer. Also for square roots, every square root gives us only positive values.