How do you evaluate #log_a x = 3/2log_a 9 + log_a 2#?

3 Answers
Feb 27, 2016

We will use the following log properties: #alogn = logn^a and log_an + log_am = log_a(n xx m)#

Explanation:

#log_ax = log_a9^(3/2) + log_a2#

#log_ax = log_asqrt(9^3) + log_a2#

#log_ax = log_a27 + log_a2#

#log_ax = log_a(27 xx 2)#

#log_ax = log_a54 -> x = 54#

Practice exercises

  1. Solve for x in the following equation. Round your answers to the nearest hundredth: #log_n2x = 1/3log_n64- 2log3#

  2. A commonly used logarithm rule is if #a = b -> loga = logb#. Solve for x in #2^(2x - 3) = 3^(3x)#

Good luck!

Feb 27, 2016

x = 54

Explanation:

using the #color(blue)" laws of logarithms "#

#• logx + logy = logxy #

#• logx - logy = log(x/y) #

#• logx^n = nlogx hArr nlogx = logx^n #

#• "If " log_b x = log_b y rArr x = y #

#rArr log_a x = log_a 9^(3/2) + log_a 2 #

[ now #9^(3/2) = (sqrt9)^3 = 3^3 = 27 #]

#rArr log_a x = log_a (27xx2) = log_a 54 #

hence x = 54

Feb 27, 2016

#x=54#

Explanation:

We are given #log_ax=3/2log_a9+log_a2#. The first thing I want to do is combine like- terms. To do that, I first must change #3/2log_a9# in to the same form as #log_a2#. That means that I just take the coefficient, #3/2#, and I raise #9# to that power, so that #3/2log_a9# becomes #log_a9^(3/2)#. I can do thanks to the third rule of logs. This says that "Logarithmic Rule 3: #log_b(m^n) = n · log_b(m)#."

Anyways, I now have my equation as #log_ax=log_a9^(3/2)+log_a2#, which can be simplified to #log_ax=log_a27+log_a2#. Now we can combine like-terms. Something you should know is that logarithms are a little backward. For example, we combine #log_a27+log_a2# by multiplying the #27# and the #2#, like this #log_a(27*2)#. This is according to the first law of #log#s.

This now gives us #log_ax=log_a54#. From here, I will rewrite this #log# into exponetial form. The format for that is as follows: #color(blue)(b)^color(red)(x)=color(green)(y)# becomes #log_color(blue)(b)color(green)(y)=color(red)(x)# and vice versa.

Applying that logic, I take #log_color(blue)(a)color(red)(x)=color(green)(log_a54)# and rewrite it as #color(blue)(cancel(a))^color(green)(cancel(log_a)54)=color(red)(x)#. Now, because #a# and #log_a# are inverses of each other, they cancel out, which brings down #color(green)(54)#, leaving us with #color(green)(54)=color(red)(x)#. There we go, we are done! Nice job.