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Chapter 3: Problem 36

Solve the exponential equation algebraically. Approximate the result to three decimal places. $$2^{x+1}=e^{1-x}$$

Short Answer

Expert verified
The exact value of x is \(\frac{1 - ln(2)}{ln(2) + 1}\). When approximated to three decimal places, the solution to the equation is \(x = 0.158\).

Step by step solution

01

Apply logarithm on both sides of the equation

First apply the natural logarithm (ln) on both sides of the equation \(2^{x+1}=e^{1-x}\). This gives us \(ln(2^{x+1})=ln(e^{1-x})\). Using the property of logarithm \(ln(a^b) = b * ln(a)\), we get \((x+1)*ln(2)=1-x\).
02

Simplify the equation

We transform the equation into the standard form \((x+1)ln(2) + x = 1\). Then distribute the ln(2) on the left side, which gives \(xln(2) + ln(2) + x = 1\). Combine like terms on the left to get \(x(ln(2) + 1) = 1 - ln(2)\).
03

Solve for x

We isolate x by dividing both sides by \(ln(2) + 1\). This gives us \(x = \frac{1 - ln(2)}{ln(2) + 1}\).

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Most popular questions from this chapter

Rewrite each verbal statement as an equation. Then decide whether the statement is true or false. Justify your answer. The logarithm of the quotient of two numbers is equal to the difference of the logarithms of the numbers.

Use a graphing utility to graph and solve the equation. Approximate the result to three decimal places. Verify your result algebraically. $$2 \ln (x+3)=3$$

Function \(\quad\) Value $$ g(x)=-\ln x \quad x=\frac{1}{2}$$

Find the domain, \(x\) -intercept, and vertical asymptote of the logarithmic function and sketch its graph. $$h(x)=\ln (x+5)$$

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