## anonymous one year ago Which logarithm is equivalent to log3^16 – log3^2? log332 log314 log38 log318

1. anonymous

2log₃(x) - 3log₃(y) + log₃(8) Get rid of coefficients: log₃(x²) + log₃(y⁻³) + log₃(8) Now just put it all together by multiplying (since now I have all +'s, notice the NEGATIVE exponent for the y logarithm): log₃(x² * y⁻³ * 8) = log₃(8x² / y³) #17 log₂(64) = ? You need to ask the question (anytime you have a logarithm), the base to the WHAT equals the argument...for instance here the base is 2 and the argument is 64, so: 2 to the WHAT equals 64? The what is what the logarithm evaluates to (exactly like saying 42/6 --> 6 * what = 42, answer 7, therefore 42/6 = 7) So 2*2 = 4, nope 2*2*2 = 8, nope...but what 8² = 64. If 8 = 2*2*2 and 8² = 64, then (2*2*2) * (2*2*2) = 64 = 2⁶ Therefore log₂(64) = 6 Edit: Incidentally, there's no way to "evaluate" a logarithm other than by guess and check...unless you use other numerical techniques...so you need to understand that when you are asked to evaluate a logarithm it will ALWAYS be special. You usually make an educated guess, then verify. For instance, instead of realizing that 8² = 64, we might have just kept multiplying by 2: 2*2 = 4 2*2*2 = 8 2*2*2*2 = 16 2*2*2*2*2 = 32 2*2*2*2*2*2 = 64 1 2 3 4 5 6 --> 2⁶ = 64 Other than that, the only other trick you need to keep in mind is when you have negative answers for the exponent: log₂(1/8) = ? 1/8 = 1/2³ = 2⁻³ Therefore log₂(1/8) = -3 ALSO remember that log(1) = 0, no matter what the base Because any base raised to 0 will ALWAYS equal 1

2. anonymous

I think its the 2nd one but i'm not sure

3. anonymous

I don't get it

4. anonymous

wait, I'm thinking this is the problem$\log_3 16 - \log_3 2$

5. ganeshie8

Ahh that makes more sense !

6. anonymous

when you combine subtracted logs, divide the arguments $\log_3 a-\log_3b=\log_3 \frac{ a }{ b }$

7. anonymous

ok

8. anonymous

so you need to divide 16 by 2

9. anonymous

that would equal 8

10. anonymous

yes

11. anonymous

so the answer would be the 3rd one

12. anonymous

looks so. pls give the question the once over when you paste it in. As you can see it causes a lot of confusion trying to guess at the question. You can use the equation tool to make it clearer