Question

Solve the following to find the value of $x$:
${4^x} + {6^x} = {9^x}$

Answer 1

Hint:- To solve this question we have to make some changes so that it can be solved so we will divide by ${9^x}$. Now make the quadratic equation by assuming like terms as one variable, solve this quadratic equation and again substitute that variable to what we had taken earlier. Use log at the end.

Complete step-by-step answer:

Here we have
${4^x} + {6^x} = {9^x}$
Now on dividing by ${9^x}$ we get,
$\dfrac{{{4^x}}}{{{9^x}}} + \dfrac{{{6^x}}}{{{9^x}}} = \dfrac{{{9^x}}}{{{9^x}}} \\\ \Rightarrow {\left( {\dfrac{4}{9}} \right)^x} + {\left( {\dfrac{6}{9}} \right)^x} = 1 \\\$
We can write it as
${\left( {{{\left( {\dfrac{2}{3}} \right)}^2}} \right)^x} + {\left( {\dfrac{2}{3}} \right)^x} = 1$
${\left( {\dfrac{2}{3}} \right)^{2x}} + {\left( {\dfrac{2}{3}} \right)^x} = 1$
Now to solve it easily let ${\left( {\dfrac{2}{3}} \right)^x} = t$ , now equation becomes quadratic in t
${t^2} + t - 1 = 0 \\\ \\\$
Using formula for finding roots of quadratic equation, we get,
$t = \dfrac{{ - 1 \pm \sqrt 5 }}{2}$
Now we have to replace t by ${\left( {\dfrac{2}{3}} \right)^x}$
${\left( {\dfrac{2}{3}} \right)^x} = \dfrac{{ - 1 \pm \sqrt 5 }}{2}$
From here to find x we have to take a log with base $\left( {\dfrac{2}{3}} \right)$ to find x.
${\log _{\dfrac{2}{3}}}{\left( {\dfrac{2}{3}} \right)^x} = {\log _{\dfrac{2}{3}}}\dfrac{{ - 1 + \sqrt 5 }}{2} \\\ \therefore x = {\log _{\dfrac{2}{3}}}\left( {\dfrac{{ - 1 + \sqrt 5 }}{2}} \right) \\\$ $\left( {\because {{\log }_a}{a^x} = x} \right)$

Note: Whenever we get this type of question the key concept of solving is we have to find x by any means one method is discussed above and there are many more methods but we have to select the shortest one and I think this method is too easy to understand.