Question

2^x+1=3^1-x then find the value of x by using logarithms​

Answer 1

AnsWer :

0.226294.

Solution :

We have, Equation.

$\tt \dagger \: \: \: \: \: {2}^{x + 1} = {3}^{1 - x}$

$\tt : \implies {2}^{x + 1} = {3}^{1 - x}$

$\tt : \implies {2}^{x} \times {2}^{1} = {3}^{1} \times {3}^{ - x}$

$\tt : \implies {2}^{x } \times {2}^{1} = \dfrac{{3}^{1}}{ {3}^{x} }$

$\tt : \implies {2}^{x } \times {3}^{x} = \dfrac{{3}^{1} }{{2}^{1} }$

$\tt : \implies {2}^{x } \times {3}^{x} = \dfrac{3}{2 }$

$\tt : \implies {6}^{x } = \dfrac{3 }{2}$

Taking Log on Both sides, We get.

$\tt : \implies log 6^x = log \dfrac{3}{2}$

$\tt : \implies x = log_{6} \dfrac{3}{2}$

$\tt : \implies x = 0.226294.$