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 logs.

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.