lisa123
  • lisa123
solve 5^x=3^2x-1
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
lisa123
  • lisa123
I know I have to take the ln of each side
Nerdsarecool
  • Nerdsarecool
What are we supposed to solve
Nerdsarecool
  • Nerdsarecool
Oh

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More answers

Nerdsarecool
  • Nerdsarecool
We need to know the question to answer it
Nerdsarecool
  • Nerdsarecool
Lets go through some simple steps
anonymous
  • anonymous
I'll watch.
Nerdsarecool
  • Nerdsarecool
So what steps to you know about this problem
lisa123
  • lisa123
solve for x
Nerdsarecool
  • Nerdsarecool
I know
Nerdsarecool
  • Nerdsarecool
Ok
Nerdsarecool
  • Nerdsarecool
So first we need to do
Nerdsarecool
  • Nerdsarecool
2x x-1=log (3 ) 5
Nerdsarecool
  • Nerdsarecool
The answer is 2.738
lisa123
  • lisa123
lol okay so 1) take ln of each side ln5x=ln 3(2x-1) 2) move exponents to the front xln(5)=2x-1ln(3) 3) expand left side xln(5)=2ln(3) +xln(3)-ln(3) 4) factor out x x[ln(5) -ln(3)]
Nerdsarecool
  • Nerdsarecool
The answer is 2.738
lisa123
  • lisa123
no the answer needs to be in terms of ln
Nerdsarecool
  • Nerdsarecool
Oh oh ooooooooooooh
Nerdsarecool
  • Nerdsarecool
I still know what it is
Nerdsarecool
  • Nerdsarecool
Its
Nerdsarecool
  • Nerdsarecool
1 Attachment
Nerdsarecool
  • Nerdsarecool
That is the answer just click that link
lisa123
  • lisa123
the answer is ln3/2ln3-ln5
Nerdsarecool
  • Nerdsarecool
Yes
lisa123
  • lisa123
how do u get that?
lisa123
  • lisa123
help me please @Zarkon
denisaboichuk
  • denisaboichuk
ln3/2ln3-ln5
lisa123
  • lisa123
is that what you got @denisaboichuk
lisa123
  • lisa123
help me plz @Nnesha
Nnesha
  • Nnesha
what do you need help with? i mean from where to start
Nnesha
  • Nnesha
\(\color{blue}{\text{Originally Posted by}}\) @lisa123 the answer is ln3/2ln3-ln5 \(\color{blue}{\text{End of Quote}}\) do you know wanna know how that iis the answer ?? or??
Nerdsarecool
  • Nerdsarecool
I already helped her
Nerdsarecool
  • Nerdsarecool
Look at the chat
lisa123
  • lisa123
yes
Nnesha
  • Nnesha
\(\color{blue}{\text{Originally Posted by}}\) @lisa123 lol okay so 1) take ln of each side ln5x=ln 3(2x-1) 2) move exponents to the front xln(5)=2x-1ln(3) 3) expand left side xln(5)=2ln(3) +xln(3)-ln(3) 4) factor out x x[ln(5) -ln(3)] \(\color{blue}{\text{End of Quote}}\) there is mistake or typo i guess in 3rd step
lisa123
  • lisa123
I meant to say expand the right side
Nnesha
  • Nnesha
\[\large\rm xln(5)=(2x-1)\ln(3)\] (2x-1) is the exponent so you should distribute it by ln(3) so ln(3) [(2x-1)] = ??
lisa123
  • lisa123
I don't know
Nnesha
  • Nnesha
familiar with the distributive property \[\rm a(b+c)=a*b+a*c=ab+ac\]multiply both terms of the parentheses by outside term
lisa123
  • lisa123
2xln 3 -ln3
Nnesha
  • Nnesha
yep correct so xln(5)=2xln(3)-ln(3) now get all x terms on one side of the equal sign
lisa123
  • lisa123
xln 5 -2xln 3- ln3
lisa123
  • lisa123
no xln5-2x ln3 +ln 3
Nnesha
  • Nnesha
hmm keep the ln(3) at the right side so -2xln(3)+xln(5)=-ln(3) now you can take out the common factor
Nnesha
  • Nnesha
just think about simple algebra question 3x=4x-5 we should move the 4x term to the left side right in order to solve for x ?
lisa123
  • lisa123
oh so x(ln 5 -2 ln3)= ln 3 divide both sides x=ln3/2ln3-ln5
Nnesha
  • Nnesha
yep right but.. what did you do to change positive ln 5 to negative and negative -2ln(3) to positive ?
lisa123
  • lisa123
Cause thats how the answer choice I have is written but how come the 2xln 3 doesn't separate into 2ln3+xln3
Nnesha
  • Nnesha
it's not `xln(3)` remember \[\rm (2x-1)\ln(3) = 2x*\ln(3) -1 *\ln(3)\] so there is not x with `1 *ln(3)`
Nnesha
  • Nnesha
and also when u got \[\rm -2xln(3)+\ln(5)=-\ln(3)\] there is a negative sign at front of 2 so take out that as well with the x varaible

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