How do you evaluate #log_5 5 = log_5 25 - log_2 8#?

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Answer 1 · 2018-03-05T10:23:17

Please see the Explanation.

By Definition, #log_ba=m hArr b^m=a#.

Now, # (1): 5^1=5 rArr log_5 5=1#.

# (2) : 5^2=25 rArr log_5 25=2#.

# (3) : 2^3=8 rArr log_2 8=3#.

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

Answer 2 · 2018-03-05T10:30:52

False statement.

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.