Question

Find x if $ {4^x} + {6^x} = {9^x} $
 $ \left( a \right)\dfrac{{\ln \left( {\sqrt 3 - 1} \right) + \ln 2}}{{\ln 2 - \ln 3}} $
 $ \left( b \right)\dfrac{{\ln \left( {\sqrt 5 - 1} \right) + \ln 2}}{{\ln 2 - \ln 3}} $
 $ \left( c \right)\dfrac{{\ln \left( {\sqrt 5 - 1} \right) - \ln 2}}{{\ln 2 - \ln 3}} $
 $ \left( d \right)\dfrac{{\ln \left( {\sqrt 5 - 1} \right) + \ln 2}}{{\ln 3 - \ln 2}} $

Answer 1

Hint: In this particular type of question use that ( $ 4 = {2^2},6 = 2.3{\text{ and }}9 = {3^2} $ ), so first simplify the equation using this then divide the whole equation by ( $ {3^{2x}} $ ) then substitute appropriate value of the simplified equation to other variable so that equation convert into a quadratic equation then apply quadratic formula and use the properties of log so use these concepts to reach the solution of the question.

Complete step-by-step answer:
Given equation:
 $ {4^x} + {6^x} = {9^x} $
Now as we know that $ 4 = {2^2},6 = 2.3{\text{ and }}9 = {3^2} $ so the above equation is written as,
 $ \Rightarrow {\left( {{2^2}} \right)^x} + {\left( {2.3} \right)^x} = {\left( {{3^2}} \right)^x} $
Now simplify we have,
 $ \Rightarrow {2^{2x}} + {2^x}{.3^x} = {3^{2x}} $
 $ \Rightarrow {2^{2x}} + {2^x}{.3^x} - {3^{2x}} = 0 $
Now divide by $ {3^{2x}} $ throughout we have,
 $ \Rightarrow \dfrac{{{2^{2x}}}}{{{3^{2x}}}} + \dfrac{{{2^x}{{.3}^x}}}{{{3^{2x}}}} - 1 = 0 $
Now simplify we have,
 $ \Rightarrow {\left( {\dfrac{2}{3}} \right)^{2x}} + {\left( {\dfrac{2}{3}} \right)^x} - 1 = 0 $
Now let, $ {\left( {\dfrac{2}{3}} \right)^x} = y $ ................. (1), so substitute this value in above equation we have,
 $ \Rightarrow {y^2} + y - 1 = 0 $
Now the equation convert in to simple quadratic equation, so apply quadratic formula we have,
 $ \Rightarrow y = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} $ , where a = 1, b = 1 and c = -1
So substitute these values we have,
 $ \Rightarrow y = \dfrac{{ - 1 \pm \sqrt {1 - 4\left( 1 \right)\left( { - 1} \right)} }}{{2\left( 1 \right)}} $
\[ \Rightarrow y = \dfrac{{ - 1 \pm \sqrt 5 }}{2}\]
\[ \Rightarrow y = \dfrac{{ - 1 + \sqrt 5 }}{2},\dfrac{{ - 1 - \sqrt 5 }}{2}\]
Now when, \[y = \dfrac{{ - 1 - \sqrt 5 }}{2}\] so from equation (1) we have,
 $ \Rightarrow {\left( {\dfrac{2}{3}} \right)^x} = y $
 $ \Rightarrow {\left( {\dfrac{2}{3}} \right)^x} = \dfrac{{ - 1 - \sqrt 5 }}{2} $
Now take (ln) on both sides we have,

 $ \Rightarrow \ln {\left( {\dfrac{2}{3}} \right)^x} = \ln \left( {\dfrac{{ - 1 - \sqrt 5 }}{2}} \right) $
Now according to logarithmic property $ \ln \left( {\dfrac{a}{b}} \right) = \ln a - \ln b{\text{ and }}\ln {\left( {\dfrac{a}{b}} \right)^c} = c\ln \left( {\dfrac{a}{b}} \right) $ so use these properties in the above equation we have,
  $ \Rightarrow x\ln \left( {\dfrac{2}{3}} \right) = \ln \left( { - 1 - \sqrt 5 } \right) - \ln 2 $
 $ \Rightarrow x\left( {\ln 2 - \ln 3} \right) = \ln \left( { - 1 - \sqrt 5 } \right) - \ln 2 $
 $ \Rightarrow x = \dfrac{{\ln \left[ { - \left( {\sqrt 5 + 1} \right)} \right] - \ln 2}}{{\ln 2 - \ln 3}} $
Now as we know negative of ln, ln (-a) is undefined so the above value is not possible.
Now when, \[y = \dfrac{{ - 1 + \sqrt 5 }}{2}\] so from equation (1) we have,
 $ \Rightarrow {\left( {\dfrac{2}{3}} \right)^x} = y $
 $ \Rightarrow {\left( {\dfrac{2}{3}} \right)^x} = \dfrac{{ - 1 + \sqrt 5 }}{2} $
Now take (ln) on both sides we have,

 $ \Rightarrow \ln {\left( {\dfrac{2}{3}} \right)^x} = \ln \left( {\dfrac{{ - 1 + \sqrt 5 }}{2}} \right) $
Now according to logarithmic property $ \ln \left( {\dfrac{a}{b}} \right) = \ln a - \ln b{\text{ and }}\ln {\left( {\dfrac{a}{b}} \right)^c} = c\ln \left( {\dfrac{a}{b}} \right) $ so use these properties in the above equation we have,
  $ \Rightarrow x\ln \left( {\dfrac{2}{3}} \right) = \ln \left( { - 1 + \sqrt 5 } \right) - \ln 2 $
 $ \Rightarrow x\left( {\ln 2 - \ln 3} \right) = \ln \left( { - 1 + \sqrt 5 } \right) - \ln 2 $
 $ \Rightarrow x = \dfrac{{\ln \left( {\sqrt 5 - 1} \right) - \ln 2}}{{\ln 2 - \ln 3}} $
So this is the required solution of the given equation.
Hence option (C) is the correct answer.

Note: Whenever we face such types of questions the key concept we have to remember is that always recall the quadratic formula which is stated above and always recall all the properties of natural log without these properties we cannot reach to the actual answer which matched the given options.