Question

Logarithms! Wait for image :)

Answer 1

which ? 4a or 4b ?

Answer 2

Both

Answer 3

use, \(\huge \log_mn=\frac{1}{\log_nm}\)

Answer 4

u = log_3 x
1/u = log_x 3

Answer 5

using that ^ , can u solve the quadratic that you'll get ?

Answer 6

@hartnn Sorry I fell asleep.. It's 7am here already >< I think I can solve it but why is 1/u = log_x 3?

Answer 7

because:
if \(\large log_bA=y \) then \(\large b^y=A \)

so if you take log base A of both sides,

\(\large log_A(b^y)=log_A(A) \)

\(\large y\cdot log_Ab=1 \)

\(\large y=\frac{1}{log_Ab} \)

Answer 8

Could someone help with the actual steps? Can't get the answer...

Answer 9

\[\huge \log_3 x=9 \log_x 3\]Our change of base formula tells us,\[\large \color{purple}{\log_a b=\frac{\log_cb}{\log_ca}}\]Where C is whatever we want it to be.
Apply this to the RIGHT side, let's change it to a base of 3, so it matches the left side,\[\huge \log_3 x=9\frac{\log_3 3}{\log_3 x}\]

Answer 10

From here, you'll need to recall that the log a value that matches its base is simply 1. So we get 1 on the top, then we'll move that denominator over to the other side,\[\huge (\log_3 x)^2=9\]

Answer 11

okay

Answer 12

Taking the square root gives us,\[\huge \log_3 x=\pm 3\]From here, do you know how to write it as an exponential? If not there is something we can do to change it.

Answer 13

Do you mean like 3^3 = x?

Answer 14

Yes :) good.

Answer 15

How about for part b?

Answer 16

For part b, let's apply the same FIRST step we did in the other problem, change of base,\[\huge \log_3x+2=3\log_x 3\]Becomes,\[\huge \log_3x+2=3\frac{\log_33}{\log_3x}\]

Answer 17

Simplifying the log3 of 3 gives us one. Then multiplying both sides by that log in the denominator gives us,\[\huge (\log_3x)^2+2\log_3x=3\]

Answer 18

Understand that step ok? When we multiply it over, we have to multiply it to both terms, the 2 and the log.

Answer 19

yup I understand

Answer 20

If we subtract 3 from both sides, we'll see that we have a quadratic function looking thing!\[\huge (\log_3x)^2+2\log_3x-3=0\]

So maybe it can be factored.
If it's really confusing, replace log_3{x} with u so u can see what's going on.\[\large u^2+2u-3=0\]

Answer 21

And it appears this one CAN be factored, which is nice.
Do you see it? :D

Answer 22

Yup! :)

Answer 23

\[\huge (\log_3x-1)(\log_3x+3)=0\]So we have our factors.
And we can solve for our 2 values of X separately from here.

Answer 24

We'll have to be careful in the end though.
Since our problem started with an X INSIDE of a log, we can't end up with any negative values for X.

Answer 25

Okay I managed to solve it! Thanks so much :)

Answer 26

Yay good job \c:/