Question

Solve 4x = 12

Answer 1

to solve this, you need to make x the subject.
if you miltiply x by 4, what can you do to get x by itself?

Answer 2

-4?

Answer 3

no, try dividing both sides by 4

Answer 4

x=3

Answer 5

Also, try just thinking what time four equals twelve

Answer 6

yep, now see that x is by itself, which makes it the subject of the equation

Answer 7

what about 4^x=12, whats the difference in these?

Answer 8

4^x = 12 is saying 4 times itself how many times is 12. In this case it won't be a whole number.

If 12 were 64 instead for instance,

4^x = 64

4^3 = 64

because 4 times itself 4 times is 64

4*4*4 = 16*4 = 64

To solve 4^x = 12 you would need to use logarithms

log(4^x) = log(12)
x*log(4) = log(12)
x = log(12)/log(4)

Answer 9

because 4 times itself 3 times is 64**

Answer 10

makes perfect sense when you explain it thank you

Answer 11

Yup :)

Answer 12

last question i dont understand on my study guide
( Use logarithmic properties to expand each of the following expressions )

Log a^2 b^3 , how can i start on this?

Answer 13

Try using this rule\[\Large \log (a)(b) = \log a+ \log b\]
\[\Large \log (a^2)(b^3 ) = \]

Answer 14

And then use this property of logarithmic functions
\[\log(x^n) =n*\log(x)\]

Answer 15

sooo.... log(a^2)(b^3)= 2loga +3logb?

Answer 16

Yep :)

Answer 17

Very nice :)

Answer 18

now to finish the problem what would i have to do? take the log of a or 2?

Answer 19

Wait.. finish what problem?

Answer 20

If there are no values given for a or b, that is fully expanded

Answer 21

2loga +3logb

means 2 times the log of a, plus 3 times the log of b

Answer 22

Well it can be simplified backwards to what you started with

\[\Large \log(a^2)(b^3)= 2loga +3logb\]
it depends if you're being asked to expand it, or condense it.

Answer 23

ok, for condensing, what steps would you have to take?

Answer 24

for example 3logd-4logf

Answer 25

Just do everything you did before in reverse

\[\Large 3 \log d-4 \log f = \]
\[\Large \log d^3 - \log f^4\]

Now use \[\Large \log a - \log b = \log \frac{ a }{ b }\]

Answer 26

@jesusrenteria1