Question

If 2^x+1=3^x-1then find the value of x. (By using logs)​

Answer 1

Answer:

The answer is $x=\dfrac{\log 3+\log 2}{\log 3-\log 2}$

Step-by-step explanation:

Given equation is

              $2^{x+1}=3^{x-1}$

Take common log both sides

           $\Rightarrow \log 2^{x+1}=\log 3^{x-1}$

Using power rule ($\log a^n=n\log a$) of logarithm

          $\Rightarrow (x+1)\log 2=(x-1)\log 3\\\Rightarrow x\log 2-x\log 3=-\log 3-\log 2\\\Rightarrow x(\log 2-\log 3)=-(\log 3+\log 2)\\\\\Rightarrow x=\dfrac{-(\log 3+\log 2)}{\log 2-\log 3}=\dfrac{\log 3+\log 2}{\log 3-\log 2}$