Question

How do you solve \[{{3}^{2x}}={{4}^{x+1}}\] ?

Answer 1

Hint: We will solve this question by applying logarithms. First we have to apply log on both sides of the equation. After applying logs on both sides simplify the equation using logarithm formulas. By further simplifying the obtained equation we can get the value of x.

Complete step by step solution:
Let us know some logarithm formulas.
\[\log \left( xy \right)=\log \left( x \right)+\log \left( y \right)\]
\[\log \left( \dfrac{x}{y} \right)=\log \left( x \right)-\log \left( y \right)\]
\[\log \left( {{x}^{y}} \right)=y\times \log \left( x \right)\]
\[{{\log }_{x}}\left( \dfrac{1}{{{x}^{a}}} \right)=-a\]
\[{{\log }_{x}}1=0\]
We will use these formulas in solving the problem where ever needed.
Given equation is
\[{{3}^{2x}}={{4}^{x+1}}\]
Let us apply logarithm on both sides of the equation.
By applying logarithm on both sides we will get
\[\Rightarrow \log \left( {{3}^{2x}} \right)=\log \left( {{4}^{x+1}} \right)\]
We can see that both logarithms are in the form of \[\log \left( {{x}^{y}} \right)\].
So we can apply the formula
\[\log \left( {{x}^{y}} \right)=y\times \log \left( x \right)\]
By applying the formula we will get
\[\Rightarrow 2x\log \left( 3 \right)=\left( x+1 \right)\log \left( 4 \right)\]
Now we have to remove the parenthesis on RHS side.
By removing parenthesis, we will get
 \[\Rightarrow 2x\log \left( 3 \right)=x\log \left( 4 \right)+\log \left( 4 \right)\]
Now we have to rearrange the terms as x containing on one side and remaining terms on other side.
By rearranging we will get
\[\Rightarrow 2x\log \left( 3 \right)-x\log \left( 4 \right)=\log \left( 4 \right)\]
We can see x in both the terms on LHS side. So we can take x as common on LHS side of the equation.
By taking x as common we will get
\[\Rightarrow x\left( 2\log \left( 3 \right)-\log \left( 4 \right) \right)=\log \left( 4 \right)\]
Now we have to isolate x on LHS side.
To isolate x on LHS side we have to divide the equation with \[2\log \left( 3 \right)-\log \left( 4 \right)\] on both sides.
On dividing the equation with \[2\log \left( 3 \right)-\log \left( 4 \right)\] we will get
\[\Rightarrow \dfrac{x\left( 2\log \left( 3 \right)-\log \left( 4 \right) \right)}{2\log \left( 3 \right)-\log \left( 4 \right)}=\dfrac{\log \left( 4 \right)}{2\log \left( 3 \right)-\log \left( 4 \right)}\]
By simplifying above equation
\[\Rightarrow x=\dfrac{\log \left( 4 \right)}{2\log \left( 3 \right)-\log \left( 4 \right)}\]
From the above equation we can write 4 as \[{{2}^{2}}\].
After rewriting 4 we will get
\[\Rightarrow x=\dfrac{\log \left( {{2}^{2}} \right)}{2\log \left( 3 \right)-\log \left( {{2}^{2}} \right)}\]
Now we can apply the formula \[\log \left( {{x}^{y}} \right)=y\times \log \left( x \right)\]
By applying the formula we will get
\[\Rightarrow x=\dfrac{2\log \left( 2 \right)}{2\log \left( 3 \right)-2\log \left( 2 \right)}\]
So by simplifying the given equation we got \[\dfrac{2\log \left( 2 \right)}{2\log \left( 3 \right)-2\log \left( 2 \right)}\].

Note: We cannot solve it without applying logarithms because if we apply exponential formulas the equation we will get is false. So we have to apply logarithms and we have to be aware of logarithm formulas.