Please lord help, I have a test tomorrow & I do not understand this one thing. 1. 7^(2x+1)=3^(x+3) 2. 9^(x+2)=2^(5x-4) 3. 3^(4x+3)=8^(-x+2) Please lord help, I have a test tomorrow & I do not understand this one thing. 1. 7^(2x+1)=3^(x+3) 2. 9^(x+2)=2^(5x-4) 3. 3^(4x+3)=8^(-x+2) @Mathematics
You're solving for x :O So yeah. I'll repost this in the morning if nobody knows right now.
take logs of both sides
Hero, my online calculator (not wolfram) shows to use ln? and they got the right answer according to my textbook.
(2x+1)log(7) = (x+3)log(3)
\[\frac{2x+1}{x+3} = \frac{\log(3)}{\log{7}}\]
simplify right side
then continue solving for x
My textbook, shows for its answer: ln(27/7) over ln(49/3), which equals x=.48
You can use ln or logs...doesn't matter
You will get the same answer
But use ln since your book says to use it
The procedure is still the same
I'm still lost. Can you do number two?
When you take logs, whatever is in the exponent comes down from being an exponent and then all you have is just the log of the base of the exponent
I get all that, I just don't get how you actually solved to get x=.48 from that problem.
9^(x+2)=2^(5x-4) Take logs of both sides: (x+2)log(9) = (5x-4) log(2) Divide both sides by (5x-4), then by log(9): (x+2)/(5x-4) = log(2)/log(9)
I thought you wanted help with the next problem
Okay, I'll finish solving the first problem
I do but i thought you were stil concerntrating on the first. just keep going with #2
First of all, if you're trying to do this without a calculator, not a good idea.
You need a calculator to evaluate log(3) and log(7)
0.56
Yes, now multiply both sides by (x+3) again
2x+1 = .56(x+3)
distribute the .56
.56x+1.68
2x+1 = .56x + 1.6973
And the answer is .48! :D
:D
I thought the answer was supposed to be 48
Oh nevermind, lol
You got it
It wasn't that hard man
Im looking at the 2nd one, and I think i am doing something wrong..
It's the same exact procedure man
(x+2)/(5x-4) = log(2)/log(9) .31
Also, you can verify your answer by plugging x back in
K so i distribute .31 through 5x-4
and make sure both sides are the same
on the test, plug x back in and verify!
x+2=-1.24+1.55 x
That's the only way to verify that you're right
If you don't do that, you will fail
If you forget to do it, you will fail
x+2=-1.24+1.55 x x=5.89091
I can promise you that
So make sure to verify your answer tomorrow
I know, thats what I was planning on doing the whole test, just checking constantly.
Good
I just wanted you to confirm that you will do it
Textbook got 5.65, but they round weird, so 5.89 is close enough :D
Do not round answers to the end. Never ever round an answer until the end
That's rule #1 of how not to make a mistake on a math test
Textbook got 5.65, but they round weird, so 5.89 is close enough :D
Eh, my teacher is lenient and doesn't care as long as you are somewhere in the ballpark.
Textbook got 5.65, but they round weird, so 5.89 is close enough :D
Like with a calculator I don't round because I just do 2nd answer, and just input like 50 decimals :P
You're lucky. 99% of most teachers aren't so nice. But still, you want to get as close to the right answer as possible. I wouldn't make it a habit of rounding prematurely man
Don't ever take an online exam where they want the exact answer
Textbook got 5.65, but they round weird, so 5.89 is close enough :D
Do you know ANY way of memorzing formulas besides writing them a million times?
Textbook got 5.65, but they round weird, so 5.89 is close enough :D
I was thinking of just learning the formulas like before the test, and right when the test starts just write them right on the test..
Yep, you just have to know them man. That's all I can say. Most of it is just algebra anyway
What is the test on?
What chapter or section of math topics does it cover?
Textbook got 5.65, but they round weird, so 5.89 is close enough :D
Exponential functions, Logarithmic functions, Properties of Logarithms, Exponential and Logarithmic Equations, and inverses of logarithms/exponential equations
log(a-b) + log(a-b) = ?
Textbook got 5.65, but they round weird, so 5.89 is close enough :D
inverse as in y=x+2, then just replacing x for y, y for x, solving for y. x=y+2, y=x-2, easy.
Well, actually, that's a bad one because the answer to that would just be 2(log(a-b)) which if you further simplify to 2(log(a/b))
Textbook got 5.65, but they round weird, so 5.89 is close enough :D
I have no idea what that is :S
you see that they have the same base so you combine them
It's like x + x
= 2x
Textbook got 5.65, but they round weird, so 5.89 is close enough :D
oh! I know what that is, i thought you wanted me to name something lmao
except it has an extra step included
Like as if its a property or something, i was like uhhh....
Yeah , but you didn't get it immediately
No, I want you to simplify it
c'mon man, you should know what to do if someone gave you that
Textbook got 5.65, but they round weird, so 5.89 is close enough :D
You multiply them btw ;)
You can't put "oh yeah I know what that is" on the test
If they're the same base you can just add and simplify as I showed above
You're forgetting the x + x rule
Textbook got 5.65, but they round weird, so 5.89 is close enough :D
ln(x+2)+ln(3x-2)=2ln(2x), you do ln(x+2)(3x-2)=ln4x^2, 3x^2+4x-4=4x^2 x^2-4x+4=0 quadraticccc
Textbook got 5.65, but they round weird, so 5.89 is close enough :D
yeah my teacher sucks she doesnt teach us rules, she just tells us to solve.
ln(3x-2)+ln(3x-2) = ?
Textbook got 5.65, but they round weird, so 5.89 is close enough :D
(3x-2)(3x-2) 9x^2-6x-6x+4 9x^2-12x+4
Textbook got 5.65, but they round weird, so 5.89 is close enough :D
or 2ln(3x-2)
Join our real-time social learning platform and learn together with your friends!