Question

If \[{\log _{175}}5x = {\log _{343}}7x\] , then the value of \[{\log _{42}}({x^4} - 2{x^2} + 7)\] is
A) 1
B) 2
C) 3
D) 4

Answer 1

Hint:
Firstly, we will equate \[{\log _{175}}5x = {\log _{343}}7x\] to k. Then we will apply log identity. Then, on simplification we will get the value of k. We will use the value of k in the equation and then we will simplify and we will get the value of x. Then we will put the value of x in \[{\log _{42}}({x^4} - 2{x^2} + 7)\] . On simplification we will get the answer.

Complete step by step solution:
Given,
  \[{\log _{175}}5x = {\log _{343}}7x\] ….(1)
Assume that \[{\log _{175}}5x = {\log _{343}}7x = k\] ….(2)
Here we will use a logarithmic identity
If \[{\log _b}x = m\]
Then,
 \[ \Rightarrow x = {b^m}\] ….(3)
Using (3) in (2), we have
 \[ \Rightarrow {\log _{175}}5x = k\]
Using log identity here, we get
 \[ \Rightarrow 5x = {175^k}\] ….(4)
Now considering \[{\log _{343}}7x = k\]
Using log identity, we get,
 \[ \Rightarrow 7x = {343^k}\] ….(5)
Dividing \[(4)\]by \[(5)\]
 \[ \Rightarrow \dfrac{{5x}}{{7x}} = \dfrac{{{{175}^k}}}{{{{343}^k}}}\]
Eliminating \[x\] from both numerator and denominator and writing 175 as \[175 = 5 \times 5 \times 5\] and 343 as \[343 = 7 \times 7 \times 7\] .
 \[ \Rightarrow \dfrac{5}{7} = {\left( {\dfrac{{5 \times 5 \times 7}}{{7 \times 7 \times 7}}} \right)^k}\]
Eliminating 7 from both numerator and denominator of R.H.S. of the above equation.
 \[ \Rightarrow \dfrac{5}{7} = {\left( {\dfrac{{5 \times 5}}{{7 \times 7}}} \right)^k}\]
\[ \Rightarrow \dfrac{5}{7} = {\left( {{{\left( {\dfrac{5}{7}} \right)}^2}} \right)^k}\]Now we write it in the in the form of square.

Multiplying 2 and k in the power of R.H.S.
 \[ \Rightarrow \dfrac{5}{7} = {\left( {\dfrac{5}{7}} \right)^{2k}}\]
Since, bases are equal therefore powers are also equal.
Therefore,
  \[ \Rightarrow 2k = 1\]
On dividing both sides by 2 we get,
 \[ \Rightarrow k = \dfrac{1}{2}\] ….(6)
Using equation (6) in (4), we get
 \[ \Rightarrow 5x = {175^{\dfrac{1}{2}}}\]
175 can be written as \[175 = 5 \times 5 \times 7\]
 \[ \Rightarrow 5x = {(5 \times 5 \times 7)^{\dfrac{1}{2}}}\]
 \[ \Rightarrow 5x = {({5^2} \times 7)^{\dfrac{1}{2}}}\]
Since, the square root of \[{5^2} \times 7\] is \[5\sqrt 7 \] .
 \[ \Rightarrow 5x = 5\sqrt 7 \]
On dividing the equation by 5 we get,
 \[ \Rightarrow x = \sqrt 7 \] ….(7)
We have to find the value of \[{\log _{42}}({x^4} - 2{x^2} + 7)\]
Putting the value of x from equation (7) in \[{\log _{42}}({x^4} - 2{x^2} + 7)\] , we get
 \[ \Rightarrow {\log _{42}}\left( {{{\left( {\sqrt 7 } \right)}^4} - 2 \times {{\left( {\sqrt 7 } \right)}^2} + 7} \right)\]
On simplification we get,
 \[ \Rightarrow {\log _{42}}({7^2} - 2 \times 7 + 7)\]
On further Simplifying we get,
 \[ \Rightarrow {\log _{42}}(49 - 14 + 7)\]
Hence, we have,
 \[ \Rightarrow {\log _{42}}42\]
We know that \[{\log _a}a = 1\] , then
 \[ \Rightarrow {\log _{42}}42 = 1\]
Therefore,
 \[ \Rightarrow {\log _{42}}({x^4} - 2{x^2} + 7) = 1\]

Hence option A is correct.

Note:
You may get stuck or find difficulty in solving equations containing log. You can also find difficulty in applying log identities and solving powers. Logarithm is the inverse of exponential.
Some basic logarithmic identities are:
 \[{\log _a}a = 1\]
If \[{\log _b}x = m\] \[ \Rightarrow x = {b^m}\]
 \[\log {}_ba = \dfrac{{\log a}}{{\log b}}\]
 \[\log {a^b} = b\log a\]
 \[\log ab = \log a + \log b\]
 \[\log \dfrac{a}{b} = \log a - \log b\]