Big Ideas Math Algebra 2, 2014
BI
Big Ideas Math Algebra 2, 2014 View details
6. Solving Exponential and Logarithmic Equations
Continue to next subchapter

Exercise 55 Page 339

Try both methods. Which seems more efficient for you?

See solution.

Practice makes perfect
We are given the logarithmic inequality shown below. log_5 x < 2 We will show how to solve it graphically and algebraically. Then, we will discuss which method is more efficient in this case.

Solving the Inequality Graphically

To solve a logarithmic inequality graphically, we can graph both sides of the inequality to see for which x-values it gets satisfied.

As we can see, the graphs intersect when x=25. Therefore, the graph of log_5 x is below the graph of y=2 when 0

Solving the Inequality Algebraically

To solve the inequality algebraically, we first need to solve the related equation. log_5 x = 2 Let's give it a try.
log_5 x = 2
â–Ľ
Solve for x

5^(LHS)=5^(RHS)

5^(log_5 x) = 5^2

log_5(5^m)=m

x = 5^2
x =25
Therefore, x=25 is a critical value. It separates the solutions set from the rest of the x-values. Since a logarithmic function is always increasing, we can know that the values 0Conclusions

Even thought preferences may vary, for this particular case graphing is simpler and faster, since the boundary point has an integer x-value.