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.

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