Exercise 84 Page 498

Start by excluding any values of y that would result in taking the logarithm of a number less than or equal to zero.

{ y | 1/3 < y <3 }

Practice makes perfect

Before we begin, remember that the argument of a logarithm has to be greater than 0. With that in mind, let's start by excluding any values of y that would result in taking the logarithm of a number less than or equal to zero. log_8 (3y-1) < log_8 (y+5) After we determine the range of values on which the inequality is defined, we can solve the inequality using the Properties of Logarithms and combine our solution sets.

Determining Where the Inequality is Defined

For the first logarithm to be defined, 3y-1 needs to be greater than zero.

3y-1 > 0

AddIneq

LHS+1>RHS+1

3y > 1

DivIneq

.LHS /3.>.RHS /3.

x > 1/3

Similarly, for the second logarithm to be defined, y+5 needs to be greater than zero.

y+5 > 0

SubIneq

LHS-5>RHS-5

y > - 5

Let's combine the solutions above to determine the interval where the inequality is defined.

The intersection of the above intervals, which is the interval where the inequality is defined, is x > 13.

Solving the Inequality

To solve an inequality with logarithms with the same base on each side, we use the Inequality Property of Logarithmic Functions.

Inequality Property of Logarithmic Functions
Ifb > 1, then log_b x > log_b y if and only if x > y. Ifb > 1, then log_b x < log_b y if and only if x < y.

We have that b=8, which is greater than 1, so we can use the above property to finish solving the inequality. log_8 (3y-1) < log_8 (y+5) ⇕ 3y-1 < y+5 Let's solve this inequality.

3y-1 < y+5

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Solve for y

AddIneq

LHS+1

3y< y+5+1

SubIneq

LHS-y

2y< 5+1

AddTerms

Add terms

2y<6