Question

3^(log x) -2^(log x)= 2^(log x+1) -3^(log x-1)

find the value of x​

Answer 1

Step-by-step explanation:

${3}^{log \: x} - {2}^{log \: x} = {2}^{log \: x + 1} - {3}^{log \: x - 1} \\ {3}^{log \: x} + {3}^{log \: x - 1} = {2}^{log \: x + 1} + {2}^{log \: x} \\ {3}^{log \: x} + {3}^{log \: x} \times {3}^{- 1} = {2}^{log \: x} \times {2}^{1} + {2}^{log \: x} \\ {3}^{log \: x}(1 + \frac{1}{3} ) = {2}^{log \: x}(2 + 1) \\ {3}^{log \: x}( \frac{4}{3} ) = {2}^{log \: x}(3) \frac{{3}^{log \: x}}{{2}^{log \: x}} = 3 \times \frac{3}{4} \\ ( \frac{3}{2} )^{log \: x} = \frac{9}{4} \\ ( \frac{3}{2} )^{log \: x} = (\frac{3}{2} )^{2} \\ log_{10} \: x \: = 2 \\ x = {10}^{2} \\ x = 100$