Chapter 1.7, Problem 25E

To determine

To solve the equation $4e^{-2x+1}=7e^{3x}$ for x and check the answer.

Answer to Problem 25E

  $x=\frac{1-\ln \left(7\right)+\ln \left(4\right)}{5}\approx 0.088$

Explanation of Solution

Given:

The equation, $4e^{-2x+1}=7e^{3x}$ .

Concept Used:

Laws of Logs:

• Law 1:   Product property of logarithm, $\ln \left(xy\right)=\ln \left(x\right)+\ln \left(y\right)$ .
• Law2:   Exponent property of logarithm, $\ln \left(x^y\right)=y\ln \left(x\right)$ .
• Law3:   $\ln \left(\frac{1}{x}\right)=-\ln \left(x\right)$
• Law 4:   Quotient property of logarithm, $\ln \left(\frac{x}{y}\right)=\ln \left(x\right)-\ln \left(y\right)$ .
• Law5:   Same base property of logarithm, $\ln \left(e\right)=1$ .
• Law6:   $\ln \left(1\right)=0$
• A logarithmic function is defined only for positive real numbers greater than zero.

Calculation:

In order to solve the given equation $4e^{-2x+1}=7e^{3x}$ for x , first take natural log both sides and simply further by using the product property and exponent property of logarithm. Then simplify further by isolating x terms on left side by performing some basic algebraic operations as shown below,

  $\begin{array}{c}4e^{-2x+1}=7e^{3x}\\ \ln \left(4e^{-2x+1}\right)=\ln \left(7e^{3x}\right)\left(\text{Take log both sides}\right)\\ \ln \left(4\right)+\ln \left(e^{-2x+1}\right)=\ln \left(7\right)+\ln \left(e^{3x}\right)\left(\text{Law 1}\right)\\ \ln \left(4\right)+\left(-2x+1\right)\ln \left(e\right)=\ln \left(7\right)+3x\ln \left(e\right)\left(\text{Law 2}\right)\\ \ln \left(4\right)+\left(-2x+1\right)\cdot \left(1\right)=\ln \left(7\right)+3x\cdot \left(1\right)\left(\text{Law 5}\right)\\ \ln \left(4\right)-2x+1=\ln \left(7\right)+3x\\ \ln \left(4\right)-2x+1-3x=\ln \left(7\right)+3x-3x\\ \ln \left(4\right)-5x+1=\ln \left(7\right)\\ \ln \left(4\right)-5x+1-\ln \left(4\right)-1=\ln \left(7\right)-\ln \left(4\right)-1\\ -5x=\ln \left(7\right)-\ln \left(4\right)-1\\ x=\frac{\ln \left(7\right)-\ln \left(4\right)-1}{-5}\\ x=\frac{1-\ln \left(7\right)+\ln \left(4\right)}{5}\approx 0.088\end{array}$

Thus,on solving the given equation it gives $x=\frac{1-\ln \left(7\right)+\ln \left(4\right)}{5}\approx 0.088$ .

Now to check the solution, substitute $x=0.088$ in the given equation and check whether the left hand side and right hand side values are equal to no. So, it gives

  $\begin{array}{c}4e^{-2x+1}=7e^{3x}\\ 4e^{-2\left(0.088\right)+1}=7e^{3\left(0.088\right)}\\ 4e^{0.824}=7e^{0.264}\\ 9.11\approx 9.11\left(\text{True Statement}\right)\end{array}$

Since the left hand side value is equal to the right hand side value, so it verifies the solution.