McGraw Hill Glencoe Algebra 2, 2012
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McGraw Hill Glencoe Algebra 2, 2012 View details
6. Common Logarithms
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Exercise 32 Page 496

{ p | p ≥ 3.5129 }

Practice makes perfect

To solve the given exponential inequality, we will use the Inequality Property of Logarithmic Functions.

Inequality Property of Logarithmic Functions

&Ifb > 1, log_b x > log_b y if and only if x > y. &Ifb > 1, log_b x < log_b y if and only if x < y.

Using this property, we can rewrite the exponential inequality as a logarithmic inequality. 5^(p-2) ≥ 2^p ⇔ log 5^(p-2) ≥ log 2^pTo solve the above logarithmic inequality, we need to recall the Power Property of Logarithms.

Power Property of Logarithms

log_b m^p = p log_b m, where m and b are positive numbers, with b≠ 1.

Let's use it and solve the inequality!

log 5^(p-2) ≥ log 2^p

log_()(a^m)= m* log_()(a)

(p-2)log 5 ≥ p log 2
p log 5 - 2 log 5 ≥ p log 2
p log 5 - p log 2 - 2 log 5 ≥ 0
p log 5 - p log 2 ≥ 2 log 5
p (log 5 - log 2 )≥ 2 log 5

The final step to isolate p will be to divide both sides of the inequality by log 5 - log 2. Before we take this step, let's stop for a moment and think about the sign of the expression. 5>2 ⇔ log 5 > log 2 Because log 5 is greater than log 2, we know that the difference between log 5 and log 2 will be a positive number. Therefore, to divide by this value, we do not have to reverse the inequality sign.

p (log 5 - log 2 )≥ 2 log 5
p ≥ 2 log 5/log 5 - log 2
p ≥ 3.5129

Finally, we can write our answer in set-builder notation. { p | p ≥ 3.5129 }