Question

Solve:
1) 3^(8-2x)=81

2) ln 4x + ln 3x = ln 13

Answer 1

1) 3 to what power is 81? Then 8-2x must be equal to that number.

2) use the properties of logs to rewrite the LHS (Left Hand Side). then eponentiate each side and solve.

Answer 2

1) =4
2)= i have no clue

Answer 3

:(

Answer 4

1) 4 = 8-2x. solve for x.

2) ln x + ln y = ln xy use this property.

then ln xy = ln z => xy = z. solve.

Answer 5

so 1) is 2?
and 2) I'm working on

Answer 6

where did z come from? not sure what to plug in for that

Answer 7

z is arbitrary, he's just using it as an example.

Answer 8

yes

Answer 9

oh okay. so to be honest i am utterly confused with this log stuff

Answer 10

Okay. You'll get a better understanding of it as you study more. For right now you just need to know specific processes. You know that ln(xy) = ln(x) +ln(y) right?

Answer 11

yes, if i use EXACT numbers.
but what do i do with the numbers after finding the ln of them?

Answer 12

ln 3x + ln 4x = ln (3x)(4x) = ln (12x^2)

Answer 13

Well with this problem you don't need to find the ln of them.

Answer 14

As pgpilot has just shown, ln(3x)+ln(4x) =ln(12x^2)

Answer 15

You therefore know that
ln(12x^2) = ln(13)

Answer 16

so 12x^2 = 13

solve for x.

Answer 17

^ Right

Answer 18

If you see two things in the same "situation" in an equation, you can remove the rest of the stuff

So like, 3^x = 3^4 you can assume that x=4 without having to deal with logs or anything complicated like that

Answer 19

Similarly, if you see that ln(x) = ln(2) you can just assume that x is equal to 2

Answer 20

Apply the same thing to your equation
ln(12x^2) = ln(13) can be changed to
12x^2 = 13

And from there it's basic algebra.

Answer 21

bnut remember that whatever you find x is, when it goes back in to the origianl equation, the argument of log or ln must be positive.
in
ln x

x is the argument. so x > 0.

Answer 22

its basic but it comes out as a huge decimal. is positive and−1.040832999733what they want?

Answer 23

You don't need to worry about that

Answer 24

\[x = \pm \sqrt{\frac{13}{12}}\] but only the positive value is valid

Answer 25

thank you guys:)

Answer 26

Yeah. That ^

Your teacher doesn't need to see 1,000,000 digits you got out of your calculator.

Answer 27

up for anymore questions?

Answer 28

If you just apply the same principles, I'm sure it'll work for the rest of your homework.

Answer 29

yeah, put it in a new post. okay?

Answer 30

But go ahead

Answer 31

thanks again!

Answer 32

no worries!