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Chapter 10: Problem 45

Solve each equation. $$\log _{5} x=-3$$

Short Answer

Expert verified
x = \(\frac{1}{125}\)

Step by step solution

01

- Understand the Logarithmic Equation

The given equation is \(\log_{5} x = -3\). This means we are looking for the value of \(x\) that satisfies this equation.
02

- Convert Logarithm to Exponential Form

To solve the equation, convert the logarithmic form to exponential form. Recall that \(\log_{a} b = c\) is equivalent to \(a^{c} = b\). For \(\log_{5} x = -3\), it converts to \(5^{-3} = x\).
03

- Calculate the Exponential Expression

Now, compute the value of the exponential expression: \(5^{-3}\). This is equivalent to \(\frac{1}{5^{3}}\).
04

- Simplify the Expression

Simplify \(\frac{1}{5^{3}}\). Calculate \(5^{3}\) which is \(5 \times 5 \times 5 = 125\). Hence, \(\frac{1}{5^{3}} = \frac{1}{125}\).
05

- State the Solution

The value of \(x\) that satisfies the equation \(\log_{5} x = -3\) is \(\frac{1}{125}\).

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Key Concepts

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

Base and Exponent Rules
Understanding the rules governing bases and exponents is crucial for solving and simplifying logarithmic and exponential equations.

Firstly, \(a^{-n} = \frac{1}{a^{n}}\): A negative exponent means you divide 1 by the base raised to the positive exponent. Secondly, \(a^{m} \times a^{n} = a^{(m+n)}\): When you multiply like bases, you add the exponents.

Knowing these rules can greatly simplify complex problems and help you convert and reduce forms accurately. If you remember that a logarithm essentially reverses exponentiation, you can confidently solve any related problem by applying these rules.

These fundamental rules are widely applicable, making them important to understand fully.

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Most popular questions from this chapter

Solve each equation. Give exact solutions. $$ \log _{5}(12 x-8)=3 $$

Solve each equation. Use natural logarithms. Approximate solutions to three decimal places when appropriate. $$ \ln e^{-x}=\pi $$

Solve each equation. Give exact solutions. $$ \log (6 x+1)=\log 3 $$

Solve each equation. Give exact solutions. $$ \log _{2} x+\log _{2}(x-6)=4 $$

Why is \(\log _{a} a=1\) true for any value of \(a\) that is allowed as the base of a logarithm?
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