Question

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

Answer 1

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!

Answer 2

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

Answer 3

`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.