Question

The original problem is: f(x)=1/2log(2x) I have to find the inverse. I switched the x's and y's. Do I need to now multiple or divide somewhere? @phi

Answer 1

original is
\[ y= \frac{1}{2} \log(2x) \]
swap x and y:
\[ x= \frac{1}{2} \log(2y) \]
ok so far ?

Answer 2

I renamed f(x) as y. (easier to type).

Answer 3

you want to solve for y. the first step is multiply both sides by 2.
can you do that ?

Answer 4

2x=log(4y)?

Answer 5

no 2x=log(2y)

Answer 6

yes 2x = log(2y) (the 2*1/2 =1 on the right side)

Answer 7

now you see the (ugly) log(2y) on the right side.
How do we get rid of log (assumed to be base 10) ?
we use a "base 10" for both sides.
can you do that ?

Answer 8

\[ 2x= 2\cdot \frac{1}{2} \log(2y) \\ 2x= \frac{2}{2} \log(2y) \\
2x= \log(2y)
\]

Answer 9

just like the previous problem except instead of "e" we use 10 as the base...

Answer 10

Do I divide the sides by 2? And where does the ten come from?

Answer 11

no, we do not divide by 2
we have
\[ 2x= \log(2y) \]
or, switching the order (which we are allowed to do)
\[ \log(2y)= 2x \]

the next step is "get at the 2y"
i.e. "eliminate the log"

to get rid of a log (which is short for \( \log_{10} \) ), we use exponentiation, which is a fancy word for make each side the exponent (tiny number in the upper right)
of a *base*
the *base* for a log is always told to you....

if you see \( \log_{n} \) then the n (it will be a number ) is the base
if you see ln, the base is "e" (a number about equal to 2.718281828...)
if you see log with not number below it, the base is assumed to be 10

Answer 12

so for out problem, what is the base ?

Answer 13

2?

Answer 14

please carefully re-read the bottom part of my last post.

Answer 15

Sorry 10

Answer 16

yes, the base is 10

we have
\[ \log(2y)= 2x \]
can you make each side the exponent of base 10 ?
in other words, write a (big) 10, under and to the left, so we have 10 to a power on both sides.

Answer 17

Would the left be 10^2x?

Answer 18

for \[ \log(2y) = 2x \]
the *right side* becomes 10^2x

Answer 19

what about the left side ?

Answer 20

I don't know

Answer 21

two parts: first can you write it with 10 as the base ?

Answer 22

10log^2x?

Answer 23

almost. the whole log(2x) becomes the exponent. try again

Answer 24

10^log2x

Answer 25

yes (except we have a y , not an x)
now you have to *memorize* that
\[ base^{\log(stuff)}= stuff \]
in other words, we can simplify
\[ 10^{\log(2y)} \] to what ? just the "stuff" inside the log!

Answer 26

So the stuff inside the log (2y) becomes the left side

Answer 27

yes. and the "stuff" is just 2y (the base and the log parts go away)

Answer 28

\[ \log(2y) = 2x \\ 10^{\log(2y)}= 10^{2x} \\ 2y = 10^{2x} \]

Answer 29

the last step is multiply both sides by 1/2

Answer 30

you get
\[ y = \frac{1}{2} 10^{2x} \]
notice you can't divide 2 into 10 because order of operations says we do exponents before divide or multiply. in other words we have to do (10^2x) (and whatever it is , we don't know how to divide it by 2, so we just leave the 1/2 out front:
\[ y = \frac{1}{2} 10^{2x} \]

Answer 31

I see, thank you

Answer 32

for more background, see Khan's videos
https://www.khanacademy.org/math/algebra2/logarithms-tutorial/logarithm_basics/v/logarithms

he has a few, and if you have time, watch them. They may help you.