Question

If the following relations are given as \[x=\log 20\] , \[y=\log \dfrac{8}{3}\] and \[z=2\log \dfrac{\sqrt{3}}{2}\] , find :
(a) \[\left( x-y-z \right)\]
(b) Prove that \[{{5}^{{{\left( x-y-z \right)}^{2}}}}=5\]

Answer 1

Hint: In part (a), we have to find the value of \[\left( x-y-z \right)\] . Use the formula, \[m\log n=\log {{n}^{m}}\] and transform the value of z. Now, put the values of x, y, and z in \[\left( x-y-z \right)\] . We know the property of logarithm, \[\log m-\log n=\log \dfrac{m}{n}\] . Use this property and simplify the equation \[\left( x-y-z \right)\] . The base of log is 10. Now, use the property, \[{{\log }_{a}}a=1\] and get the value of \[\left( x-y-z \right)\] . In part (b) we have to prove \[{{5}^{{{\left( x-y-z \right)}^{2}}}}=5\] . In the LHS, we have \[{{5}^{{{\left( x-y-z \right)}^{2}}}}\] . Now, put the value of \[\left( x-y-z \right)\] and get the value of the LHS.

Complete step by step solution:
According to the question, we have the values of x, y, and z.
\[x=\log 20\] …………………..(1)
\[y=\log \dfrac{8}{3}\] …………………….(2)
\[z=2\log \dfrac{\sqrt{3}}{2}\] ……………………………(3)
We know the formula, \[m\log n=\log {{n}^{m}}\] ………………………..(4)
Now, from equation (3) and equation (4), we get
\[\Rightarrow z=\log {{\left( \dfrac{\sqrt{3}}{2} \right)}^{2}}\]
\[\Rightarrow z=\log \left( \dfrac{3}{4} \right)\] ………………………..…(5)
In part (a), we have to get the value of \[\left( x-y-z \right)\] ……………………..…..(6)
From equation (1), equation (2), and equation (5), we have the values of x, y, and z.
Now, putting the values of x, y, and z in equation (6), we get
\[=\left( x-y-z \right)\]
\[=\log 20-\log \dfrac{8}{3}-\log \left( \dfrac{3}{4} \right)\] …………………………………..(7)
We know the property of logarithm, \[\log m-\log n=\log \dfrac{m}{n}\] ………………………….(8)
Now, from equation (7) and equation (8), we get
\[=\log 20-\log \dfrac{8}{3}-\log \left( \dfrac{3}{4} \right)\]
\[=\log \left( \dfrac{20}{\dfrac{8}{3}} \right)-\log \left( \dfrac{3}{4} \right)\]
\[=\log \left( \dfrac{60}{8} \right)-\log \left( \dfrac{3}{4} \right)\] …………………………….(9)
Simplifying equation (9) by using the property shown in equation (8), we get
\[=\log \left( \dfrac{60}{8} \right)-\log \left( \dfrac{3}{4} \right)\]
\[\begin{align}
  & =\log \left( \dfrac{\left( \dfrac{60}{8} \right)}{\left( \dfrac{3}{4} \right)} \right) \\
 & =\log \left( \dfrac{60\times 4}{8\times 3} \right) \\
 & =\log \left( \dfrac{240}{24} \right) \\
\end{align}\]
\[=\log 10\] ……………………(10)
We know the property, \[{{\log }_{a}}a=1\] ………………………….(11)
The base of log is 10.
Now, transforming equation (11), we get
\[=log10={{\log }_{10}}10\] ……………………………….(12)
Using the property shown in equation (11) and simplifying equation (12), we get
\[{{\log }_{10}}10=1\] .
Therefore, the value of \[\left( x-y-z \right)\] is 1 ………………….(13)
In part (b), we have to prove the equation \[{{5}^{{{\left( x-y-z \right)}^{2}}}}=5\] .
In the LHS of the above equation, we have \[{{5}^{{{\left( x-y-z \right)}^{2}}}}\] ………………………….(14)
The RHS of the equation is equal to 5 ……………………………………(15)
From equation (13), we have the value of \[\left( x-y-z \right)\] equal to 1.
Now putting the value \[\left( x-y-z \right)\] in equation (14), we get
\[{{5}^{{{\left( x-y-z \right)}^{2}}}}={{5}^{{{1}^{2}}}}={{5}^{1}}=5\] ……………………….(16)
After solving the LHS, we get 5 and from equation (15) we have the RHS equal to 5. So,
LHS = RHS.
Hence, proved.

Note: In this question, one might take the base of log equal to e and then get the value of \[{{\log }_{e}}10\] . This is wrong because the base of log is equal to 10. So, we have to keep in mind that the base of log is equal to 10. Care must be taken while applying the logarithmic properties. While simplifying the expression (x-y-z), we must take the corresponding terms properly and apply the property. This is another step where students tend to make silly mistakes.