Without a calculator?

Question

does anyone know the answer to evaluating log
___1_
7 ^49

and can anyone evaluate (27)^2/3

Answer 1

As written (log of [1/7 to the 49th power]) you can simplify the first one, and you can calculate it without calculating 7^49, but without at least knowing the log of 7, I'm not sure how you'd calculate it out:

log(1/(7^49)) = log(7^-49) = -49 * log(7).

The subsequent responder's guess, that you mean log-base-7(1/49) is more reasonable and can be done without a calculator.

The trick to evaluating the second one is to know that 27 = 3*3*3 = 3^3:

27^(2/3) = (3^3)^(2/3) = 3^(3 * 2/3) = 3^2 = 9

This assumes that you want 27 raised to the (2/3) power, and not "27 squared, divided by three."

Answer 2

I'm not really sure if you want the decimal value for the first question so just simplified it:
1) the problem can be written in this form:
log (7^(-49))
2) in logs, you can take the exponent and put it in the front:
-49 log 7

second question. cant tell if its (27)^(2/3) or (27^2)/3
for (27)^(2/3) :
1) this basically means, the cubed root of 27. then square it.
2) cubed root of 27 is 3
3) then take the value 3 and square it
4) answer is 9

for (27^2)/3
1) split the 27^2 into 27 times 27
2) (27 times 27) / 3
3) separate them --> (27/3)(27)
4) (9)(27) = 243

Answer 3

do you mean log base 7 of 1/49

If so, that means what exponent for 7 will give me 1/49
well 7^2 = 49 and make it negative to do reciprocal 1/49

ANSWER = -2

check 7^(-2) = 1/49

=]


If you meant log base 49 of 7.
Then what exponet for 49 gives 7
ANSWER=1/2

Please let me know if you meant something else, I can do it also
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(27)^2/3

exponent 2/3 means cube root(1/3) and then square(2)

cube root of 27 = 3
3 squared = 9

(27)^2/3 = 9

Answer 4

I/m guessing that you mean 1/(7^49) so it's
7^(-49) and
log(7^(-49)) = -49*log(7) = -41.4098

27^(2/3) = (cube root 27)² = (3)² = 9

HTH


Doug