Evaluate logs

[Guest]

Evaluate:

log subscript 3 [27 * (81^(1/3)] + log subscript 5(125 * 5^(1/4)

I need help figuring this one out:
I just know: log subscript 2[3^3 * ( 3^(4/3)] log subscript 5(5^3 * 5^(1/4)

what are the rules from here that are suppose to be used?
thanks for the help

 

[stapel]

Is the first log "base 3" or "base 2"? (You've written it both ways, is why I ask.)

For the second log, naturally you'd want to combine the argument into five to some power, since log₅(5^a) = a log₅(5) = a.

Thank you.

Eliz.

 

[Guest]

oo sry base 3, typo the answer is 77/12 though and I keep getting small faction numbers..>.<

 

[Gene]

log₃[27 * (81^(1/3)] + log₅(125 * 5^(1/4)
You mix rules here (I think)
log₃[3^3 * ( 3^(4/3)] log₅(5^3 * 5^(1/4)
It is true that log(5)+log(3)=log(5*3) but not that log(5)+log(3)=log(5)*log(3)

From what Eliz told you
log₃[3^3]+log₃[3^(4/3)] + log₅[5^3]+log₅5^(1/4) =
3 + 4/3 + 3 + 1/4 =
(36+16+36+3)/(3*4) = 91/12
Either you have another typo or a wrong answer.

 

[Guest]

yep 91/12 is correct our teacher just told us that today..back of the text book had the wrong answer..what a surprise =p haha

 

[soroban]

Hello, bittersweet!

Evaluate: \(\displaystyle \log_3\left(27\cdot81^{\frac{1}{3}}\right)\,+\,\log_5\left(125\cdot5^{\frac{1}{4}}\right)\)

You started correctly . . . now keep going.

The first log is: \(\displaystyle \log_3\left(3^3\cdot3^{\frac{4}{3}}\right)\;=\;\log_3\left(3^{\frac{13}{3}}\right)\;=\;\frac{13}{3}\cdot\log_3(3)\;=\;\frac{13}{3}\cdot1\;=\;\frac{13}{3}\)

The second log is \(\displaystyle \,\log_5\left(5^3\cdot5^{\frac{1}{4}}\right)\;=\;\log_5\left(5^{\frac{13}{4}}\right)\;=\;\frac{13}{4}\cdot\log_5(5)\;=\;\frac{13}{4}\cdot1\;=\;\frac{13}{4}\)

Therefore, the answer is: \(\displaystyle \L\,\frac{13}{3}\,+\,\frac{13}{4}\;=\;\frac{91}{12}\)