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log(subscrip)3 (square root of) 243

2007-04-23 03:31:14 · 4 answers · asked by Anonymous in Science & Mathematics Mathematics

4 answers

The square root comes out directly: 1/2[log3 3 +log3 81]
which can be simplified even more to equal: 1/2[1+4]=5/2

2007-04-23 03:36:40 · answer #1 · answered by bruinfan 7 · 0 0

log[base 3]( sqrt(243) )

To solve this without a calculator, first, express sqrt(243) as 243 to the power of (1/2).

log[base 3] ( (243)^(1/2) )

As per the log identities, move the (1/2) in front of the log.

(1/2) log[base 3](243)

243 = 3^5, so we have

(1/2) log[base 3](3^5)

And we can move the 5 outside of the log.

(1/2)(5) log[base 3](3)

log[base 3](3) is just equal to 1, so we have

(1/2)(5)(1)

(5/2)

2007-04-23 10:39:09 · answer #2 · answered by Puggy 7 · 0 0

log [base 3] √243 =
log [base 3] (243)^(1/2) =
(1/2) log [base 3] (243) =
(1/2) log [base 3] (243) =
(1/2) log [base 3] (81*3) =
(1/2) log [base 3] (9*9*3) =
(1/2) log [base 3] (3*3 * 3*3 * 3) =
(1/2) log [base 3] (3)^5 =
(5/2) log [base 3] (3)=
(5/2) * 1 = 5/2

2007-04-23 10:37:21 · answer #3 · answered by Anonymous · 0 0

y = log3 [sqrt(243)]
y = 1/2*log3(243)
2y = log3(243)
3^(2y) = 243

We know that 3^5 = 243
So y = 2.5

2007-04-23 10:45:34 · answer #4 · answered by Dr D 7 · 0 0

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