Evaluating logarithmic expressions...quite simple, but not for me?

Question

evaluate:
log base 3 of 3 4th root of [3]

ln 1

Answer 1

Assuming this is "log base 3 of [3 times the fourth root of 3]"...

You need to know two things to solve this problem:

1) The x'th root of y is the same as y^(1/x). So, for example, the fourth root of 81 is 81^(1/4).

2) log-base-a(a^n) = n. That's the very definition of log-base-a.

So...

log3(3 * 3^(1/4)) =
log3(3^(5/4)) =
5/4 =
1.25

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Second problem:

ln(1) = x

Is another way of saying...

e^x = 1

What number do you have to raise e to, in order to get 1? the answer is 0. Any number to the 0th power is equal to 1. Thus:

ln(1) = 0

Answer 2

Ok, a nice way to set it up is...

3^x = 3 times the fourth root of 3.

So, to do this take the log of both sides...

log3^x = log (3 times fourth root of 3)

You can rewrite log3^x as xlog3 by some rule...forgot which.

xlog3 = log (3 times fourth root of 3)

Divide both sides by log 3...

x = log (3 times fourth root of 3) / log 3 ->
x = .5964 / .4771

x = 1.25

To check..

3^x = 3 times the fourth root of 3.

3 ^ x = 3.948

3 ^ 1.25 = 3.948

So log base 3 of 3 4th root of [3] = 1.25


And the ln 1 is 0. When doing ln, think of it in terms of e. So when evaluating ln1, think "e up to what power (x) = 1?" Any number up to the 0 power is 1, so e ^ 0 = 1, meaning the ln of 1 is 0.

Answer 3

If you mean log[base 3] 3^34, then the answer is 34.

When he base of you log and the base of your exponent are the same, then the answer is the value of the exponent. In this case, the base of the log is 3 and the base of your exponent is 3 as well, so your answer is 34 which is the exponent.
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If you mean log[base 3] (3*3^4), then you answer is 5.

log (a*b) = log a + log b
log[base 3] (3*3^4) = log[base 3] 3 + log[base 3] 3^4

Using the explanation given above,
log[base 3] 3 = 1 and log[base 3] 3^4 = 4

1+4 = 5
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ln 1 = 0

ln = log[base e]

log[base e] 1 = log[base e] e^0 = 0

This follows from the same explanation given for the first question