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Chapter 9: Problem 81

Solve. If no solution exists, state this. $$ \log _{5} \sqrt{x^{2}-9}=1 $$

Short Answer

Expert verified
\( x = \pm \sqrt{34} \)

Step by step solution

01

Rewrite the Logarithmic Equation in Exponential Form

The given logarithmic equation is \( \log_{5}(\sqrt{x^2 - 9}) = 1\). Rewrite this equation in its exponential form. Recall that \( \log_b(y) = c \) can be rewritten as \( b^c = y \).So, \( 5^1 = \sqrt{x^2 - 9} \). Simplify to get: \( 5 = \sqrt{x^2 - 9} \).
02

Remove the Square Root

To eliminate the square root, square both sides of the equation.\( 5^2 = (\sqrt{x^2 - 9})^2 \) becomes \( 25 = x^2 - 9 \).
03

Isolate the \( x^2 \) Term

Add 9 to both sides of the equation to isolate the \( x^2 \) term:\( 25 + 9 = x^2 - 9 + 9 \) simplifies to \( 34 = x^2 \).
04

Solve for \( x \)

Take the square root of both sides to solve for \( x \):\( x = \pm \sqrt{34} \). Since \(\sqrt{34}\) is irrational, the solution in radical form is \( x = \pm \sqrt{34} \).
05

Verify the Solutions

Substitute \( x = \sqrt{34} \) and \( x = -\sqrt{34} \) into the original logarithmic equation to verify: \( \sqrt{(\sqrt{34})^2 - 9} \) yields \( \sqrt{34 - 9} = \sqrt{25} = 5 \) and \( \log_5(5) = 1 \). Both solutions, \( x = \sqrt{34} \) and \( x = -\sqrt{34} \), satisfy the original equation.

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Key Concepts

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

Logarithms
Logarithms are a way to express exponential relationships. The equation \(\text{log}_b(y) = c\) means that \(b\) raised to the power of \(c\) equals \(y\). For instance, \(\text{log}_2(8) = 3\) because \(2^3 = 8\). Logarithms help in solving equations where the unknown variable is an exponent.
Logarithms have two important properties:
  • The Product Rule: \(\text{log}_b(xy) = \text{log}_b(x) + \text{log}_b(y)\)
  • The Quotient Rule: \(\text{log}_b\bigg(\frac{x}{y}\bigg) = \text{log}_b(x) - \text{log}_b(y)\)
In our exercise, we used the property \(\text{log}_5(\text{something}) = 1\) which translates to \(5^1 = \text{something}\).
Exponential Form
Rewriting logarithmic equations in exponential form makes them easier to solve. Exponential form shows the relationship directly. For example, \(\text{log}_5(\text{something}) = 1\) becomes \(5^1 = \text{something}\).
In our exercise, we started with \(\text{log}_5(\text{something}) = 1\) and rewrote it in exponential form as \(5 = \text{something}\). This step simplified finding the solution.
Square Roots
Square roots help reverse the squaring of numbers. The square root of a number \(x\) is a value that, when multiplied by itself, gives \(x\). For instance, \(\text{The square root of 25 is both 5 and -5 because } 5^2 = 25 \text{ and } (-5)^2 = 25\).
In our exercise, we had \(\text{5} = \text{鈭�(x虏鈭�9)}\). We squared both sides to remove the square root, turning it into \(\text{25} = \text{(x虏鈭�9)}\).
  • Remember, squaring both sides always makes sure you check your answers, as squaring can introduce extraneous solutions.
Radical Form
Radical form is a way to express roots, such as square roots. For a number \(x\), the square root is written as \(\text{鈭歺}\). Radicals can simplify representations, especially when perfect squares are involved.
In our exercise, after finding that \(x虏 = 34\), we took the square root to get \(x = 卤鈭�34\).
  • \(x = 卤 5.831\) is the decimal form, whereas \(x = 卤鈭�34\) is the radical form.
Always check whether the radical form simplifies further to help in understanding the solution better.

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