Question

How do you solve ${{e}^{7x}}=-9{{e}^{2x}}$ ?

Answer 1

Hint: To solve ${{e}^{7x}}=-9{{e}^{2x}}$ , we have to take natural logarithm on both sides. Then use the logarithm rules $\ln \left( ab \right)=\ln a+\ln b$ , $\ln \left( {{a}^{m}} \right)=m\ln a$ and $\ln \left( e \right)=1$ to make the equation in the form $5x=\ln \left( -9 \right)$ . We will then use $\ln \left( -9 \right)=\ln \left( -1\times 9 \right)$ in the previous equation to make the equation in the form $5x=\ln \left( -1\times 9 \right)$ . We will then use logarithmic rules and the formula $\ln \left( -1 \right)=i\pi $ , to solve this equation further and get the value of x.

Complete step by step solution:
We have to solve ${{e}^{7x}}=-9{{e}^{2x}}$ . Let us take natural logarithms on both sides. The given equation becomes
$\ln \left( {{e}^{7x}} \right)=\ln \left( -9{{e}^{2x}} \right)$
We know that $\ln \left( ab \right)=\ln a+\ln b$ . Hence, the above equation becomes
$\ln \left( {{e}^{7x}} \right)=\ln \left( -9 \right)+\ln \left( {{e}^{2x}} \right)$
We know that $\ln \left( {{a}^{m}} \right)=m\ln a$ . Therefore, we can write the above equation as
$7x\ln \left( e \right)=\ln \left( -9 \right)+2x\ln \left( e \right)$
We know that $\ln \left( e \right)=1$ . Hence, the above equation can be written as
$\begin{align}
  & 7x.1=\ln \left( -9 \right)+2x.1 \\
 & \Rightarrow 7x=\ln \left( -9 \right)+2x \\
\end{align}$
Let us move 2x to the RHS. We will get
$\begin{align}
  & \Rightarrow 7x-2x=\ln \left( -9 \right) \\
 & \Rightarrow 5x=\ln \left( -9 \right) \\
\end{align}$
We know that $-9=-1\times 9$ . Therefore, we can write the above equation as
$\Rightarrow 5x=\ln \left( -1\times 9 \right)$
Let us apply the rule $\ln \left( ab \right)=\ln a+\ln b$ in the above equation.
$\Rightarrow 5x=\ln \left( -1 \right)+\ln 9$
We know that $\ln \left( -1 \right)=i\pi $ . Hence, we can write the above equation as
$\Rightarrow 5x=i\pi +\ln 9$
Let us move 5 from LHS to RHS.
$\begin{align}
  & \Rightarrow x=\dfrac{i\pi +\ln 9}{5} \\
 & \Rightarrow x=\dfrac{i\pi }{5}+\dfrac{\ln 9}{5} \\
\end{align}$
We know that $\ln 9=2.197$ and $\pi =3.14$ . Hence, the above equation becomes
$\Rightarrow x=\dfrac{i\times 3.14}{5}+\dfrac{2.197}{5}$
Let us solve the above equation.
$\Rightarrow x=0.439+0.628i$
Hence, the answer is $x=0.439+0.628i$ .

Note: We must note that logarithmic values for negative number is undefined. But we can find the value as a complex number. If we never consider the complex value, there is no solution to the given question.