/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 45 Solve the exponential equation a... [FREE SOLUTION] | 91影视

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Chapter 3: Problem 45

Solve the exponential equation algebraically. Approximate the result to three decimal places. $$2^{3-x}=565$$

Short Answer

Expert verified
The solution to the exponential equation is approximately \( x = -6.136 \).

Step by step solution

01

Application of logarithm

First, apply the logarithm to both sides to get rid of the base of the exponential term. Logarithms and exponents are inverses of each other, so they can counteract each other. By applying logs, we get the following equation: \( \log_2(2^(3-x)) = \log_2(565) \). The log base 2 on the left side cancels out the base 2 on the exponent, resulting in \( 3-x = \log_2(565) \).
02

Isolate the variable

Next, isolate the variable x by moving the 3 to the right side of the equation. You can do this by subtracting 3 from both sides. As a result, the equation becomes: \( -x = \log_2(565) - 3 \).
03

Solve for x

Finally, multiply the equation by -1 in order to solve for x and get it by itself on one side of the equation. We get \( x = -\log_2(565) + 3 \). Now we can evaluate the logarithm and subtract it from 3 to find the approximate value of x. Using a scientific calculator, the \(\log_2(565)\) is approximately 9.135709286, so we get \( x = -9.135709286 + 3 = -6.135709286 \).
04

Rounds

Now let's round the result to three decimal places. The result is \( x = -6.136 \).

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Most popular questions from this chapter

A cell site is a site where electronic communications equipment is placed in a cellular network for the use of mobile phones. The numbers \(y\) of cell sites from 1985 through 2008 can be modeled by \(y=\frac{237,101}{1+1950 e^{-0.355 t}}\) where \(t\) represents the year, with \(t=5\) corresponding to \(1985 .\) (Source: CTIA-The Wireless Association) (a) Use the model to find the numbers of cell sites in the years 1985,2000 , and 2006 . (b) Use a graphing utility to graph the function. (c) Use the graph to determine the year in which the number of cell sites will reach 235,000 . (d) Confirm your answer to part (c) algebraically.

Use the following information for determining sound intensity. The level of sound \(\boldsymbol{\beta}\), in decibels, with an intensity of \(I\), is given by \(\boldsymbol{\beta}=10 \log \left(I / I_{0}\right),\) where \(I_{0}\) is an intensity of \(10^{-12}\) watt per square meter, corresponding roughly to the faintest sound that can be heard by the human ear. In Exercises 65 and 66 , find the level of sound \(\boldsymbol{\beta}\). Due to the installation of noise suppression materials, the noise level in an auditorium was reduced from 93 to 80 decibels. Find the percent decrease in the intensity level of the noise as a result of the installation of these materials.

(a) solve for \(P\) and (b) solve for \(t\). $$A=P\left(1+\frac{r}{n}\right)^{n t}$$

Use the following information for determining sound intensity. The level of sound \(\boldsymbol{\beta}\), in decibels, with an intensity of \(I\), is given by \(\boldsymbol{\beta}=10 \log \left(I / I_{0}\right),\) where \(I_{0}\) is an intensity of \(10^{-12}\) watt per square meter, corresponding roughly to the faintest sound that can be heard by the human ear. In Exercises 65 and 66 , find the level of sound \(\boldsymbol{\beta}\). Due to the installation of a muffler, the noise level of an engine was reduced from 88 to 72 decibels. Find the percent decrease in the intensity level of the noise as a result of the installation of the muffler.

Use the Richter scale \(R=\log \frac{l}{I_{0}}\) for measuring the magnitudes of earthquakes. Find the magnitude \(R\) of each earthquake of intensity \(I\) (let \(I_{0}=1\) ). (a) \(I=199,500,000\) (b) \(I=48,275,000\) (c) \(I=17,000\)
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