How do you evaluate log_5 5 = log_5 25 - log_2 8log55=log525−log28?

2 Answers
Ratnaker Mehta
Mar 5, 2018

Please see the Explanation.

Explanation:

By Definition, log_ba=m hArr b^m=alogba=m⇔bm=a.

Now, (1): 5^1=5 rArr log_5 5=1(1):51=5⇒log55=1.

(2) : 5^2=25 rArr log_5 25=2(2):52=25⇒log525=2.

(3) : 2^3=8 rArr log_2 8=3(3):23=8⇒log28=3.

:. log_2 8-log_5 25=3-2=1=log_5 5.

Somebody N.
Mar 5, 2018

False statement.

Explanation:

First:

25=5^2

8=2^3

Substituting these in our equation.

log_(5)5=log_(5)5^2-log_(2)2^3

By the laws of logarithms:

log_(b)a^c=clog_(b)a

So:

log_(5)5=log_(5)5^2-log_(2)2^3= log_(5)5=2log_(5)5-3log_(2)2

Also by the laws of logarithms:

log_(a)a=1

The logarithm of the base is always 1bb

Then:

log_(5)5=2log_(5)5-3log_(2)2

log_(5)5=2-3=-1

But:

log_(5)5=1

The statement is false.

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