## anonymous one year ago Solve 3(2x) = 7(x−1).

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1. nincompoop

what do you think you should do first?

2. anonymous

its x=7

3. 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

4. nincompoop

5. anonymous

but if I do the substitution method wouldn't I need to do something else first? @nincompoop

6. nincompoop

substitution method how?

7. anonymous

its more like 3^(2x) = 7^ (x-1)

8. nincompoop

ohhh

9. nincompoop

hahaha

10. nincompoop

did you learn about logarithms yet?

11. anonymous

yes but it makes no sense

12. nincompoop

laughing out loud

13. anonymous

this is serious lol

14. anonymous

I need this class

15. anonymous

Yeah nincompoop.

16. nincompoop

use this $$\large log(a^b) = b~log(a)$$

17. nincompoop

|dw:1437603054834:dw|

18. anonymous

so I multiply them?

19. nincompoop

no do the log first

20. anonymous

so the logs are in front of the numbers?

21. nincompoop

ye

22. nincompoop

23. anonymous

will you still help me

24. nincompoop

maybe, but read and take notes first

25. anonymous

okay

26. 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. ?

27. anonymous

omgoodness thank you so much

28. 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?