log x = 0.24

find log 100x^2

We will use logarithm properties to solve the problem.

Let y = log 100x^2

We know that:

log a*b = log a + log b

==> log 100x^2 = log 100 + log x^2

Now let us rewrtie:

100 = 10^2

==> log 100x^2 = log 10^2 + log x^2

Now we know that :

log a^b = b*log a

==> log 100x^2 = 2log10 + 2log x

We know that log 10 = 1

==> log 100x^2 = 2*1 + 2*0.24

= 2+ 0.48

= 2.48

**==> log 100x^2 = 2.48**

logx = 0.24.

To find log100x^2.

Solution:

log100x^2 = log100 +logx^2, as logab = loga+logb by property of logarithms.

log100 x ^2 = log100 +2logx , as loga^m = m loga by property og logarithms.

log100 = log10 (10^2) = 2 , as loga (a^2) = 2 by proerty of logarithms.

Therefore sobstituting log100 = 2 and logx = 0.24 in log 100x^2 = log 100 + 2lofx, we get:

log100x^2 = 2+2(0.24) = 2.48