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Chapter 13: Problem 44

Solve each logarithmic equation. $$\log _{6}(5 y+1)=2$$

Short Answer

Expert verified
To solve the logarithmic equation \(\log_{6}(5y + 1) = 2\), first convert to exponential form: \(6^{\log_{6}(5y+1)} = 6^2\). Then, simplify the equation: \(5y + 1 = 36\). Isolate the variable y: \(5y = 35\). Finally, determine the value of y: \(y = \frac{35}{5}\), which gives \(y = 7\).

Step by step solution

01

Convert to exponential form

To convert the equation from logarithmic to exponential form, we use the rule \(a^{\log_a(x)}=x\). In our case, the base is 6: $$6^{\log_{6}(5y+1)} = 6^2$$
02

Simplify the equation

Now, we can simplify our equation, and we get: $$5y + 1 = 6^2$$
03

Continue simplifying

Now, let's calculate the value of \(6^2\), which is equal to 36: $$5y + 1 = 36$$
04

Isolate the variable y

To isolate the variable y, we will first subtract 1 from both sides of the equation: $$5y = 36 - 1$$ Now, we have: $$5y = 35$$ Now, we will divide both sides by 5 to find the value of y: $$y = \frac{35}{5}$$
05

Determine the value of y

By dividing 35 by 5, we obtain the value of y: $$y=7$$ Thus, the solution to the given logarithmic equation is \(y = 7\).

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Key Concepts

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

Exponential Form
To solve a logarithmic equation, the first step is often to convert it to an exponential form. This transformation makes equations easier to manipulate. Logarithms and exponents are two sides of the same mathematical coin. They help us manage numbers of different scales.
For example, in the exercise, we have the equation \(\log_{6}(5y+1) = 2\). This tells us what exponent 6 must be raised to in order to get \(5y+1\).
Using the exponential rule \(a^{\log_a(x)} = x\), we switch from the logarithmic form to exponential form:
  • Here, the base \(a\) is 6.
  • The left side becomes \(6^{\log_6(5y+1)}\), which simplifies directly to \(5y+1\).
  • So the equation converts to \(5y+1 = 6^2\).
The exponential form helps us see the direct relationships between numbers.
Solving Equations
Solving equations involves systematically working out the value of the unknown variable. Transforming the equation into an easily solvable form is the key. After converting to exponential form and simplifying, what follows is basic arithmetic.
The task now is to perform operations to isolate the variable, from our equation \(5y + 1 = 36\). Here is the step-by-step process:
  • Calculate \(6^2\), which equals 36.
  • Substitute back to get \(5y + 1 = 36\).
  • Solve for \(5y\) by subtracting 1 from both sides.
  • Finally, divide the equation \(5y = 35\) by 5 to find \(y\).
At each step, we simplify the equation further, bringing us closer to finding the solution.
Base of Logarithms
The base of a logarithm is an essential component that defines how the logarithm operates. It decides what the number is raised to in the logarithmic expression. In solving logarithmic equations, understanding the base is crucial.
In our present case, the logarithm has the base 6, \(\log_6\). The equation can be translated into how many times 6 must multiply itself to reach another number.
Key points about logarithm bases:
  • A logarithm like \(\log_6(x)\) is asking: **To what power must 6 be raised to get \(x\)?**
  • The base helps in determining the exponential form as seen in \(a^{\log_a(x)}\).
  • Misunderstanding the base can lead to incorrect solutions, so it's crucial to identify it right away.
Knowing and applying the base of logarithms correctly makes the equation much more manageable.
Isolate the Variable
Isolating the variable is often the final and most crucial step in solving equations. This involves getting the unknown variable by itself on one side of the equation, enabling us to solve for its value.
In our case, after simplifying to \(5y = 35\), isolating \(y\) involves a few simple arithmetic operations:
  • First, subtract any constants added to the variable from both sides of the equation.
  • Next, divide by the coefficient of the variable to get it alone.
Thus, with \(y = \frac{35}{5} = 7\), we successfully isolated \(y\). This let us find the specific value that satisfies the equation. Being systematic about isolating variables ensures precise and accurate solutions in various contexts of mathematics.

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Most popular questions from this chapter

The number of bacteria, \(N(t),\) in a culture \(t\) hr after the bacteria is placed in a dish is given by $$N(t)=5000 e^{00617 t}$$ a) How many bacteria were originally in the culture? b) How many bacteria are present after 8 hr?

Use the formula \(A=P\left(1+\frac{r}{n}\right)^{n t}\) to solve each problem. How much will Anna owe at the end of 4 yr if she borrows \(\$ 5000\) at a rate of \(7.2 \%\) compounded weekly?

Radioactive carbon- 14 is a substance found in all living organisms. After the organism dies, the carbon- 14 decays according to the equation $$y=y_{0} e^{-0.000121 t}$$ where \(t\) is in years, \(y_{0}\) is the initial amount present at time \(t=0,\) and \(y\) is the amount present after \(t\) yr. a) If a sample initially contains 15 g of carbon- 14 how many grams will be present after 2000 yr? b) How long would it take for the initial amount to decay to 10 g? c) What is the half-life of carbon- \(14 ?\)

Write as a single logarithm. Assume the variables are defined so that the variable expressions are positive and so that the bases are positive real numbers not equal to \(1 .\) $$\log _{6} y-\log _{6} 3-3 \log _{6} z$$

Write as a single logarithm. Assume the variables are defined so that the variable expressions are positive and so that the bases are positive real numbers not equal to \(1 .\) $$\log _{7} d-\log _{7} 3$$
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