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Chapter 3: Problem 55

Writing a Natural Logarithmic Equation In Exercises \(53-56,\) write the exponential equation in logarithmic form. $$e^{-0.9}=0.406 \ldots$

Short Answer

Expert verified
The logarithmic form of the equation \(e^{-0.9}=0.406 \ldots\) is \( ln(0.406 \ldots)=-0.9 \)

Step by step solution

01

- Understand the Exponential Equation

Given an exponential equation \(e^{-0.9}=0.406 \ldots\). Here base is \( e \), exponent is \( -0.9 \) and the result is \( 0.406 \ldots\)
02

- Converting to Logarithmic Form

In exponential form \(b^{x}=y\), corresponds to logarithmic form as \( \text{log}_{b}(y)=x. \) So here, the given equation in logarithmic form will be \( \text{log}_{e}(0.406 \ldots)=-0.9 \)
03

- Write the final logarithmic form

There is a special notation for logarithms with base \(e \), it's written as ln. The equation in logarithmic form is \( ln(0.406 \ldots)=-0.9. \)

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Most popular questions from this chapter

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

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

Writing a Natural Logarithmic Equation In Exercises \(53-56,\) write the exponential equation in logarithmic form. $$e^{-0.9}=0.406 \ldots$

Solve the logarithmic equation algebraically. Approximate the result to three decimal places. $$\ln (x+1)-\ln (x-2)=\ln x$$

Solve the equation algebraically. Round your result to three decimal places. Verify your answer using a graphing utility. $$\frac{1+\ln x}{2}=0$$
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