anonymous
  • anonymous
Solve 3(2x) = 7(x−1).
Algebra
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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nincompoop
  • nincompoop
what do you think you should do first?
anonymous
  • anonymous
its x=7
nincompoop
  • nincompoop
you can deal with distributive property first or you can settle the coefficients by making the right hand side and left hand side have the same value of coefficients

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nincompoop
  • nincompoop
please do not give out answers
anonymous
  • anonymous
but if I do the substitution method wouldn't I need to do something else first? @nincompoop
nincompoop
  • nincompoop
substitution method how?
anonymous
  • anonymous
its more like 3^(2x) = 7^ (x-1)
nincompoop
  • nincompoop
ohhh
nincompoop
  • nincompoop
hahaha
nincompoop
  • nincompoop
did you learn about logarithms yet?
anonymous
  • anonymous
yes but it makes no sense
nincompoop
  • nincompoop
laughing out loud
anonymous
  • anonymous
this is serious lol
anonymous
  • anonymous
I need this class
anonymous
  • anonymous
Yeah nincompoop.
nincompoop
  • nincompoop
use this \(\large log(a^b) = b~log(a) \)
nincompoop
  • nincompoop
|dw:1437603054834:dw|
anonymous
  • anonymous
so I multiply them?
nincompoop
  • nincompoop
no do the log first
anonymous
  • anonymous
so the logs are in front of the numbers?
nincompoop
  • nincompoop
ye
nincompoop
  • nincompoop
please read this page then proceed with your problem later http://www.mathsisfun.com/algebra/exponents-logarithms.html
anonymous
  • anonymous
will you still help me
nincompoop
  • nincompoop
maybe, but read and take notes first
anonymous
  • anonymous
okay
freckles
  • freckles
Here is an example: \[5^x=6^{2x+3} \\ \text{ take } \log( ) \text{ of both sides } \\ \log(5^x)=\log(6^{2x+3}) \\ \text{ we do this so we can use } \log(r^x)=x \log(r) \text{ this is called power rule } \\ \text{ so bringing the powers down } \\ x \log(5)=(2x+3) \log(6) \\ \text{ distribute on the right hand side } \\ x \log(5)=2x \log(6)+3 \log(6) \\ \text{ now collect your terms with } x \text{ on one side } \\ x \log(5)-2x \log(6)=3 \log(6) \text{ I just subtracted } 2x \log(6) \text{ on both sides } \\ \text{ now we are going to factor the } x \text{ out on the left } \\ x(\log(5)-2 \log(6))=3 \log(6) \\ \text{ now we are going to divide both sides by what } x \\ \text{ is being multiplied by } \\ x=\frac{3 \log(6)}{\log(5)-2 \log(6) }\] Can you show what you have doing just this step: so you try taking the log of both sides and bringing down the powers. ?
anonymous
  • anonymous
omgoodness thank you so much
freckles
  • freckles
Did you want to show me the one step I was mentioning? Like taking log of both sides then bringing down the powers as a result of using the power rule for log?

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