How do you solve 4^(p-1) ≤ 3^p4p−1≤3p?

1 Answer
Daniel L.
Apr 3, 2016

The answer is : p<=log_{4/3}(3)-1p≤log43(3)−1

Explanation:

We start from the inequality :

4^{p-1}<=3^p4p−1≤3p

If we rewrite the right side as 3^{p-1}*33p−1⋅3 we will get the same exponents on both sides:

4^{p-1}<=3*3^{p-1}4p−1≤3⋅3p−1

Now if we divide both sides of the inequality by: 3^{p-1}3p−1 we will get the exponents only on the right side:

(4/3)^{p-1}<=3(43)p−1≤3

Now we can write the right side as the exponent using rule which says that: a=b^{log_a b}a=blogab

We get then:

(4/3)^{p-1}<=(4/3)^(log_{4/3}(3))(43)p−1≤(43)log43(3)

Now, when we have powers with the same bases we can write our inequality as inequality of the exponents:

p-1<=log_{4/3}(3)p−1≤log43(3) (*)

Finally we can move 11 to the right side to get the solution

p<=log_{4/3}(3)+1p≤log43(3)+1

Note:

In the expression marked with (*) if the base was lower than 1 we would have to change the sign of the inequality fron <=≤ to >=≥ when changing from exponential inequality to inequality of exponents.

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