Solve 3(2x) = 7(x−1).

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Solve 3(2x) = 7(x−1).

Algebra
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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what do you think you should do first?
its x=7
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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Other answers:

please do not give out answers
but if I do the substitution method wouldn't I need to do something else first? @nincompoop
substitution method how?
its more like 3^(2x) = 7^ (x-1)
ohhh
hahaha
did you learn about logarithms yet?
yes but it makes no sense
laughing out loud
this is serious lol
I need this class
Yeah nincompoop.
use this \(\large log(a^b) = b~log(a) \)
|dw:1437603054834:dw|
so I multiply them?
no do the log first
so the logs are in front of the numbers?
ye
please read this page then proceed with your problem later http://www.mathsisfun.com/algebra/exponents-logarithms.html
will you still help me
maybe, but read and take notes first
okay
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. ?
omgoodness thank you so much
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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