Question

If \[x = \log 0.6\], \[y = \log 1.25\] and \[z = \log 3 - 2\log 2\] then the value of \[p = {5^{x + y - z}}\] is equal to
(A) 7
(B) 0
(C) 1
(D) 2

Answer 1

Hint: We solve the equation of p by taking logarithms on both side of the equation and using the property of logarithm \[d\log c = \log {c^d}\] to break the equation into simpler parts, and substitute the values of x, y and z from the question in the equation. Then we solve the equation formed by using arithmetic properties of log which is \[\log a + \log b = \log (ab)\] and\[\log a - \log b = \log (\dfrac{a}{b})\]. At last we take antilog on both sides to obtain the value of p.

Complete Step-by-step Solution
Let us assume, \[p = {5^{x + y - z}}\] …(1)
Take the logarithm on both sides of the equation (1) we get,
\[ \Rightarrow \log p = \log ({5^{x + y - z}})\]
Using the formula \[d\log c = \log {c^d}\]we can write
\[ \Rightarrow \log p = (x + y - z)\log 5\]
Now put the value of \[x = \log 0.6\], \[y = \log 1.25\] and \[z = \log 3 - 2\log 2\] in above equation
\[ \Rightarrow \log p = [\log 0.6 + \log 1.25 - (\log 3 - 2\log 2)]\log 5\]\[ \Rightarrow \log p = [\log 0.6 + \log 1.25 - \log 3 + \log {2^2}]\log 5\]
In above equation convert multiplication of number and logarithm number by using formula \[\log a + \log b = \log (ab)\]
\[ \Rightarrow \log p = \left[ {\log (0.6 \times 1.25 \times 4) - \log (3)} \right](\log 5)\]
Now we use the formula \[\log a - \log b = \log (\dfrac{a}{b})\]
\[ \Rightarrow \log p = \left[ {\log \left( {\dfrac{{0.6 \times 1.25 \times 4}}{3}} \right)} \right](\log 5)\]
Solve the value inside the bracket by adding the terms in the numerator and then dividing the numerator by denominator.
\[ \Rightarrow \dfrac{{0.6 \times 1.25 \times 4}}{3} = \dfrac{3}{3} = 1\]
Therefore, we get the RHS of the equation as
\[ \Rightarrow \log p = \left[ {\log 1} \right](\log 5)\]
We know that the value of \[\log 1\] is 0. So, we get,
\[ \Rightarrow \log p = 0 \times \log 5\]
\[ \Rightarrow \log p = 0\] … (2)
Now do the antilog on both sides of the equation (2)
\[ \Rightarrow {e^{\log p}} = {e^0}\]
Since we know any number having power zero becomes equal to 1
\[ \Rightarrow {e^{\log p}} = 1\]
Since, log and antilog cancel each other.
\[ \Rightarrow p = 1\]
Substituting the value of p from equation (1)
\[ \Rightarrow {5^{x + y - z}} = 1\]

$\therefore $The required value of \[ {5^{x + y - z}} = 1\]. Hence, option C is the correct option.

Note:
Students are likely to make calculations very complex when they try to solve the question without using properties of logarithm, students are advised to break log terms as much as possible to make calculations easier. As in answer we first do the log on both sides after that solving the equation don’t forget to do antilog of that. Mostly students forgot to do this antilog operation and get errors in answers.