How do you solve #4^(p-1) ≤ 3^p#?

1 Answer
Daniel L.
Apr 3, 2016

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

Explanation:

We start from the inequality :

#4^{p-1}<=3^p#

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

#4^{p-1}<=3*3^{p-1}#

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

#(4/3)^{p-1}<=3#

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

We get then:

#(4/3)^{p-1}<=(4/3)^(log_{4/3}(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)# (*)

Finally we can move #1# to the right side to get the solution

#p<=log_{4/3}(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.

Related questions
See all questions in Logarithmic Models
Impact of this question
1556 views around the world
You can reuse this answer
Creative Commons License