Question

How do you solve \[{{2}^{x+1}}={{5}^{1-2x}}\] ?

Answer 1

Hint: In the given question, we have been asked to solve the exponential expression. In order to solve the given exponential expression there is only a way that is to pass both the terms as arguments of the logarithmic functions in base e. Then using the property of logarithmic i.e. \[\log {{\left( x \right)}^{a}}=a\log x\] , we will simplify the expression. Later using the distributive property of multiplication we will simplify the expression further and solve for the value of ‘x’. in this way we will get the required value of ‘x’.
Formula used:
The property of logarithm which states that,
  \[\log {{\left( x \right)}^{a}}=a\log x\]
Distributive property of multiplication over addition and subtraction states that the product of a number say ‘a’ with product of two numbers, say ‘b’ and ‘c’ is equal to the addition or subtraction of products of the number ‘a’ multiplied separately with ‘b’ and ‘c’:
  \[a\times \left( b\pm c \right)=ab\pm ac\]

Complete step by step solution:
We have given that,
  \[\Rightarrow {{2}^{x+1}}={{5}^{1-2x}}\]
Taking logarithmic function both the side of the equation, we will obtain
  \[\Rightarrow \ln {{2}^{x+1}}=\ln {{5}^{1-2x}}\]
Using the property of logarithm which states that \[\log {{\left( x \right)}^{a}}=a\log x\]
We will obtain,
  \[\Rightarrow \left( x+1 \right)\ln 2=\left( 1-2x \right)\ln 5\]
Simplifying the above using the distributive property of multiplication i.e. \[a\times \left( b\pm c \right)=ab\pm ac\] ,
  \[\Rightarrow x\ln 2+\ln 2=\ln 5-2x\ln 5\]
Rewrite the above equation as,
  \[\Rightarrow x\ln 2+2x\ln 5=\ln 5-\ln 2\]
Taking ‘x’ from LHS of the equation,
  \[\Rightarrow x\left( \ln 2+2\ln 5 \right)=\ln 5-\ln 2\]
Dividing both the sides of the equation by \[\left( \ln 2+2\ln 5 \right)\] , we will get
  \[\Rightarrow x=\dfrac{\ln 5-\ln 2}{\ln 2+2\ln 5}\]
Hence, this is the required answer.
So, the correct answer is “ \[ x=\dfrac{\ln 5-\ln 2}{\ln 2+2\ln 5}\] ”.

Note: Here, students note a point here that the base of both the exponential terms are not same or equal so there is only 1 way to solve the given exponential expression there is only a way that is to pass both the terms as arguments of the logarithmic functions in base e.
To solve these types of questions, we used the basic formulas of logarithm. Students should always require to keep in mind all the formulae for solving the question easily. After applying log formulae to the equation, we need to solve the equation in the way we solve general linear equations. Logarithmic functions are the inverse of exponential functions with the same bases