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Chapter 12: Problem 18

Solve. Where appropriate, include approximations to three decimal places. $$ 5^{x+2}=15 $$

Short Answer

Expert verified
x is approximately -0.318

Step by step solution

01

Rewrite the Equation

Start by rewriting the equation in a more manageable form. Notice that the base 5 can be isolated by dividing both sides by 5^2:\[5^{x+2} = 15 \]Divide by 5^2:\[5^x = \frac{15}{25}\]Simplify:\[5^x = 0.6\]
02

Apply the Logarithm

Apply the natural logarithm (ln) to both sides of the equation to get rid of the exponent on the left hand side:\[\text{ln}(5^x) = \text{ln}(0.6)\]Using the properties of logarithms, this can be rewritten as:\[x \text{ln}(5) = \text{ln}(0.6)\]
03

Solve for x

Solve for x by dividing both sides by ln(5):\[x = \frac{\text{ln}(0.6)}{\text{ln}(5)}\]Now use a calculator to find the value of ln(0.6) and ln(5):\[\text{ln}(0.6) \text{ is approximately } -0.511 \]\[\text{ln}(5) \text{ is approximately } 1.609 \]Divide these values to get:\[x \text{ is approximately } \frac{-0.511}{1.609} \text{ which is approximately } -0.318\]

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Key Concepts

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

logarithms
Logarithms are one of the fundamental concepts in algebra, especially when dealing with exponential equations. They help us to solve equations where the variable is an exponent. A logarithm is the inverse of an exponentiation. For example, if you have an equation like \(b^x = y\), you can solve for \(x\) using a logarithm: \(x = \text{log}_b(y)\). This makes logarithms very powerful for simplifying complex equations.
natural logarithms
A natural logarithm is a specific type of logarithm. It uses the base \(e\), where \(e\) is an irrational number approximately equal to 2.718281828. This type of logarithm is denoted as \(\text{ln}(x)\). Natural logarithms are especially useful when dealing with continuous growth or decay, commonly seen in natural sciences. In our exercise, we use natural logarithms to isolate and solve for \(x\). By applying \(\text{ln}\) to both sides of the equation, we can leverage properties of logarithms that simplify our task.
simplifying equations
Simplifying equations is a critical step in solving most algebra problems. It involves rewriting equations to make them easier to work with. For example, dividing out constants, combining like terms, or using identities are common techniques. In this exercise, simplifying the initial exponential equation by dividing both sides by \(5^2\) helps us transform it into a more manageable form. This step-by-step simplification is key to solving complex equations more easily.
exponential functions
Exponential functions are functions where the variable appears as an exponent, like \(5^x\) in our exercise. These functions are characterized by rapid growth or decay and are commonly found in numerous disciplines, including finance and physics. To solve equations involving exponential functions, we often rely on logarithms to bring the variable down from the exponent. Understanding how exponential functions work and the properties they exhibit is vital for solving such equations effectively.

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Most popular questions from this chapter

Simplify. $$\log _{e} e^{m}$$

$$\text { If } \log _{a} x=2, \text { what is } \log _{1 / a} x ?$$

If \(\log _{b} a=x,\) does it follow that \(\log _{a} b=1 / x ?\) Why or why not?

Stellar Magnitude. The apparent stellar magnitude \(m\) of a star with received intensity \(I\) is given by $$ m(I)=-(19+2.5 \cdot \log I) $$ where \(I\) is in watts per square meter \(\left(\mathrm{W} / \mathrm{m}^{2}\right) .\) The smaller the apparent stellar magnitude, the brighter the star appears. a) The intensity of light received from the sun is \(1390 \mathrm{W} / \mathrm{m}^{2} .\) What is the apparent stellar magnitude of the sun? b) The 5 -m diameter Hale telescope on Mt. Palomar can detect a star with magnitude \(+23 .\) What is the received intensity of light from such a star?

How could you convince someone that $$\log _{a} c \neq \log _{c} a ?$$
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