## Nick88888888 one year ago Write as one logarithm with base b: 3 logb x + 2 logb y

1. Nick88888888

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2. UsukiDoll

we need our three log rules $\log_b(mn) =\log_bm+\log_bn$ $\log_b(\frac{m}{n}) =\log_bm-\log_bn$ $\log_b(m^n) =n \cdot \log_bm$

3. UsukiDoll

so for this problem we don't need the subtraction one.

4. UsukiDoll

so anything to do with addition as seen in your problem will become multiplication as the end result... but first we need to use the third rule because you got exponents

5. UsukiDoll

|dw:1437293044978:dw| so use the third rule first on this

6. UsukiDoll

use $\log_b(m^n) =n \cdot \log_bm$ which is the log rule for exponents first on your drawing

7. Nick88888888

would using Log x + Log y - Log z = Solution: Log (xy)/z work? Yea i know it doesnt have z but would this method work?

8. UsukiDoll

ummm yeah but where is the base though? like ok... let's have the base be b $Log_b X + Log_b Y - Log_bZ$ then it's a combination of the first and second log rule. Your final answer should look like $Log_b(\frac{xy}{b})$ because we have a + sign which means multiplication according to the log rules and we hae a - sign which means division according to the log rules as well.

9. UsukiDoll

OMG apologizes typing too fast $Log_b(\frac{XY}{Z})$

10. Nick88888888

Logb(3(2)/b)?

11. UsukiDoll

but you just have addition for this though so the second rule isn't necessary |dw:1437293682407:dw|

12. UsukiDoll

on top of that $\log_b(m^n) =n \cdot \log_bm$ <-- this exponent rule comes first $\log_b(mn) =\log_bm+\log_bn$ <-- this addition rule comes second

13. UsukiDoll

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14. UsukiDoll

$\log_b(m^n) =n \cdot \log_bm$ use the left side of this rule m = x n = 3 for the first m = y n = 2 for the second

15. UsukiDoll

I'll do one and you can do the other one using exponent rule for logs |dw:1437294023458:dw|

16. Nick88888888

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17. UsukiDoll

alright.... now we just need the first exponent rule... so since we have addition our final result should be multiplication |dw:1437294541977:dw| fill in the blank

18. Nick88888888

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19. UsukiDoll

whoa we do not combine exponents... but the xy part is correct.

20. UsukiDoll

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21. UsukiDoll

the exponents couldn't be combined anyway due to different variables

22. Nick88888888

so its just logb(x^3y^2)?

23. UsukiDoll

yes |dw:1437294894273:dw|

24. Nick88888888

thank you