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Chapter 4: Problem 114

Explain the process to solve the equation \(\log _{b} 5+\log _{b}(x-3)=4\)

Short Answer

Expert verified
The solution is \( x = \frac{b^4}{5} + 3 \).

Step by step solution

01

Use the Product Rule of Logarithms

Combine the logarithms on the left-hand side using the product rule of logarithms: \ \ \( \log_{b} 5 + \log_{b} (x-3) = \log_{b} [5(x-3)] \)
02

Equate the combined logarithm to 4

Rewrite the equation with the combined logarithm: \ \ \( \log_{b} [5(x-3)] = 4 \)
03

Convert the logarithmic equation to exponential form

Rewrite the equation in its exponential form to solve for \( x \): \ \ \( b^{4} = 5(x-3) \)
04

Isolate the variable x

Solve for \( x \) by isolating it on one side: \ \ \( 5(x-3) = b^4 \) \ \ Divide both sides by 5: \ \ \( x-3 = \frac{b^4}{5} \) \ \ Then, add 3 to both sides: \( x = \frac{b^4}{5} + 3 \)

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Key Concepts

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

product rule of logarithms
In the first step, we use the product rule of logarithms to simplify the equation. The product rule states that the logarithm of a product is equal to the sum of the logarithms of the factors. Specifically, \(\log_{b}(MN) = \log_{b}(M) + \log_{b}(N)\).

For the equation given, we have two logarithms added together: \(\log_{b} 5 + \log_{b} (x-3)\). By applying the product rule, the equation simplifies to \(\log_{b} [5(x-3)]\).

This step reduces the number of logarithmic terms and helps in simplifying the overall equation. Once combined, the problem becomes easier to work with.
exponential form
After simplifying with the product rule, the next step involves converting the logarithmic equation to exponential form. The equation now looks like \(\log_{b} [5(x-3)] = 4\).

Converting this to exponential form means rewriting it as \(b^{4} = 5(x-3)\). This transformation helps us move from a logarithmic equation to an algebraic equation, which is generally easier to solve.

Remember, changing from the logarithmic form \(\log_{b}(A) = C\) to the exponential form means writing \(b^{C} = A\). This step is crucial in isolating the variable in the next part of our solution.
isolating the variable
Finally, we solve for the variable by isolating it on one side of the equation. Starting from our exponential form \(b^{4} = 5(x-3)\), we need to isolate \(x\).

Here鈥檚 how we do it:
  • First, divide both sides of the equation by 5: \(x-3 = \frac{b^{4}}{5}\)
  • Next, add 3 to both sides to isolate \(x\): \(x = \frac{b^{4}}{5} + 3\)

Now, we have fully solved for \(x\). At this point, you can plug in values for \(b\) if they are given, to find the numerical value of \(x\).

Isolating the variable is often the final and critical step in solving equations, and it requires careful manipulation to ensure the variable stands alone on one side of the equation.

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Most popular questions from this chapter

Solve the equation. Write the solution set with the exact values given in terms of common or natural logarithms. Also give approximate solutions to 4 decimal places. \(21,000=63,000 e^{-0.2 t}\)

(See Example 8 ) a. Estimate the value of the logarithm between two consecutive integers. For example, \(\log _{2} 7\) is between 2 and 3 because \(2^{2}<7<2^{3}\). b. Use the change-of-base formula and a calculator to approximate the logarithm to 4 decimal places. c. Check the result by using the related exponential form. $$ \log _{3} 15 $$

Given \(f(x)=b^{x},\) then \(f^{-1}(x)=\) _____ for \(b>0\) and \(b \neq 1\).

Graph the following functions on the window [-3,3,1] by [-1,8,1] and comment on the behavior of the graphs near $$ \begin{array}{l} x=0 \\ \mathrm{Y}_{1}=e^{x} \\ \mathrm{Y}_{2}=1+x+\frac{x^{2}}{2} \\ \mathrm{Y}_{3}=1+x+\frac{x^{2}}{2}+\frac{x^{3}}{6} \end{array} $$

(See Example 8 ) a. Estimate the value of the logarithm between two consecutive integers. For example, \(\log _{2} 7\) is between 2 and 3 because \(2^{2}<7<2^{3}\). b. Use the change-of-base formula and a calculator to approximate the logarithm to 4 decimal places. c. Check the result by using the related exponential form. $$ \log _{8} 5 $$
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