f(x) = log x

g(x) = x^2

fog (10 ) = ?

First we need to determine fog (x):

We know that :

fog (x)= f (g(x)

Now we will substitute with g(x) value:

==> f(g(x) = f (x^2)

Now in f(x) we will substitute with x^2:

==> f(g(x) = log x^2

Now from logarithm properties we know that:

log a^b = b *log a

==> f(g(x)) = 2*log x

Now we will calculate fog (10)

==> fog (10 ) = f(g(10)) = 2*log 10

But log 10 = 1

==> fog (10) = 2*1 = 2

**Then fog (10) = 2**

fog(10) is the same as writing f(g(10)). Now we are given that f(x) = log x and g(x) = x^2.

To find fog(10)= f(g(10)), we first find g(10).

This is equal to 10^2.

Now substitute g(10) = 10^2 as the x value in f(x).

We get f(g(10)) = f( 10^2) = log (10^2)

Now we use the relation that log (a^b) = b* log a.

log (10^2) = 2 * log 10.

Now if the base of the log is 10 log 10= 1.

Else we can write the general result for fog(10) as 2 log 10.

f(x) = logx and g(x) - x^2.

To find fog(10):

We first find the function fog(x).

f(x) =logx.

Therefore fog(x) = log{g(x)}.

fog(x) = log x^2 .

fog(x) = 2logg x , as loga^m = mloga.

Therefore fog(10) = 2 log 10.

fog(x) = 2 , as log10 = 1.