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Chapter 4: Problem 21

Solve each exponential equation. Express irrational solutions as decimals correct to the nearest thousandth. $$0.05(1.15)^{x}=5$$

Short Answer

Expert verified
x \approx 32.960

Step by step solution

01

Isolate the exponential expression

Divide both sides of the equation by 0.05 to isolate the exponential term: \[ (1.15)^x = \frac{5}{0.05} \]
02

Simplify the right side

Calculate the division on the right side of the equation: \[ \frac{5}{0.05} = 100 \]So the equation becomes: \[ (1.15)^x = 100 \]
03

Take the natural logarithm of both sides

Apply the natural logarithm to both sides of the equation to facilitate solving for x: \[ \text{ln}\big((1.15)^x\big) = \text{ln}(100) \]
04

Use the power rule of logarithms

Use the power rule, which states \( \text{ln}(a^b) = b\text{ln}(a) \), to bring the exponent x in front of the natural logarithm: \[ x \text{ln}(1.15) = \text{ln}(100) \]
05

Solve for x

Divide both sides by \( \text{ln}(1.15) \) to solve for x: \[ x = \frac{\text{ln}(100)}{\text{ln}(1.15)} \]Calculate the values: \[ x = \frac{4.60517}{0.13976} \ x \ \approx 32.960 \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Natural Logarithm
To solve exponential equations, we often use the natural logarithm (ln). The natural logarithm is the logarithm to the base **e** (approximately 2.718). It helps us with solving equations where the variable appears in the exponent.
For example, in the equation \( (1.15)^x = 100 \), we apply the natural logarithm to both sides. This gives us:
\[ \text{ln}\big((1.15)^x\big) = \text{ln}\big(100\big) \]
Using the properties of logarithms makes it easier to solve for the variable x.
Power Rule of Logarithms
The power rule of logarithms is a key concept when dealing with logarithmic expressions. It states:
\[ \text{ln}\big(a^b\big) = b \cdot \text{ln}\big(a\big) \]
This rule is useful because it allows us to bring the exponent in front of the logarithm.
In our example \[ \text{ln}\big((1.15)^x\big) = \text{ln}(100) \], using the power rule, we get:
\[ x \cdot \text{ln}(1.15) = \text{ln}(100) \]
By applying the power rule, we've simplified the equation, making it easier to isolate x and solve for it.
Irrational Solutions
An irrational number is a number that cannot be expressed as a simple fraction. When solving certain exponential equations, we may encounter irrational solutions.
In the final step of our example, we have:
\[ x = \frac{\text{ln}(100)}{\text{ln}(1.15)}\]
This calculation might give us an irrational number. To make it practical, we can express the solution as a decimal rounded to the nearest thousandth.
Here, \[ x \approx 32.960 \]
We round it to three decimal places, providing a useful approximation for most purposes.

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Most popular questions from this chapter

Evolution of Language The number of years, \(n\), since two independently evolving languages split off from a common ancestral language is approximated by $$ n \approx-7600 \log r $$ where \(r\) is the proportion of words from the ancestral language common to both languages. (a) Find \(n\) if \(r=0.9\) (b) Find \(n\) if \(r=0.3\) (c) How many years have elapsed since the split if half of the words of the ancestral language are common to both languages?

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In calculus, it is shown that $$ e^{x}=1+x+\frac{x^{2}}{2 \cdot 1}+\frac{x^{3}}{3 \cdot 2 \cdot 1}+\frac{x^{4}}{4 \cdot 3 \cdot 2 \cdot 1}+\frac{x^{5}}{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}+\cdots $$ By using more terms, one can obtain a more accurate approximation for \(e^{x}\). Use the terms shown, and replace \(x\) with \(-0.05\) to approximate \(e^{-0.05}\) to four decimal places. Check your result with a calculator.
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