Question

GET A MEDAL AND A FAN AND HELP ME WITH LOGARITHMS

Answer 1

\[\log(ab) = \log(a) +\log(b)\]\[\log\left(\frac{a}{b}\right) = \log(a)-\log(b)\]

Answer 2

\[\log(x^n) = n\log(x)\] where n is an exponent to some degree.

Answer 3

Ok...

Answer 4

So x in this case is \[\sqrt{3x-4}\]

Answer 5

and the exponent is 2?

Answer 6

I'm working on solving it also.

Answer 7

Since all the log functions are a base of 2, I guess that's helpful.

Answer 8

Ok take your time

Answer 9

You need to be careful here.
Look at this very simple calculation:
8 + 5 - 3
You must add 8 and 5 together first, so the second line is:
13 - 3
Finally, you do the subtraction:
10

Answer 10

The example with the logs is similar.
You can use the rules you wrote above, but you must do things in the order they appear from left to right.

Answer 11

Wait a minute, what do I do with the 6 and the \[\sqrt{3x-4}\]

Answer 12

the rules involving PEMDAS apples to log functions as well.

Answer 13

This is how I would do this.
First, rewrite the square root as a power.

Answer 14

\(6 \log_2 \sqrt{3x - 4} - \log _2 \dfrac{5}{x} + \log_2 5\)

Answer 15

First, copy everything above and just rewrite \(\sqrt{3x - 4} \) as a power.
Remember that a square root is the same as the power 1/2

Answer 16

\(\large 6 \log_2 (3x - 4)^{\frac{1}{2}} - \log _2 \dfrac{5}{x} + \log_2 5\)
Ok so far?

Answer 17

Yes

Answer 18

Now use the rule that the log of a power is the power times the log on the exponent of 1/2

Answer 19

so \[\frac{ 1 }{ 2 } (3x-4)\]

Answer 20

In other words, multiply 6 by 1/2 and take the exponent off 3x - 4

Answer 21

No.
The 1/2 multiplies the log, but since the log is already being multiplied by 6, 1/2 * 6 = 3. Then you have the log of 3x - 4. Like this:
\(\large 3 \log_2 (3x - 4) - \log _2 \dfrac{5}{x} + \log_2 5\)

Answer 22

oh ok I understand that now

Answer 23

Great,
Now we use the first rule above which is when you add logs, it's the log of the product.

Answer 24

Wait.

Answer 25

To use the log rula above:
log a + log b = log ab
The logs can;t have coefficients before them.
That means we need to put the 3 now as an exponent of the 3x - 4 expression like this:

Answer 26

\(\large \log_2 (3x - 4)^3 - \log _2 \dfrac{5}{x} + \log_2 5\)

Answer 27

Now we are ready to deal with the subtraction of logs.
The subtraction of logs equals the logo of the division.
log a - log b = log (a/b)

Answer 28

\(\large \log_2 \dfrac{(3x - 4)^3}{\frac{5}{x}} + \log_2 5\)

Answer 29

Ok

Answer 30

Now we can simplify the complex fraction.

Answer 31

\(\large \log_2 \dfrac{x(3x - 4)^3}{5} + \log_2 5\)

Answer 32

Now we are adding two logs.
The sum of logs is the log of the product.

Answer 33

Wait, why the the x move into the numerator?

Answer 34

b/(a/c) = bc/a

Answer 35

Oh ok thank you