We have to find x so that log (a) x= log (a) 7 + log(a) 3 - log(a) 11

Now we know that log a + log b = log ab and log a-log b = log(a/b)

log (a) x= log (a) 7 + log(a) 3 - log(a) 11

=> log(a) x = log(a) 7*3/11

=> log(a) x = log(a) (21/11)

Taking the antilog of both the sides we get x = 21/11.

**Therefore x = 21/11**

Q: What is x if loga x= loga 7+loga 3-loga 11:

Ans:

We know loga m+loga n = loga m*n and loga m - loga n = loga (m/n). We use these properties to solve the problem.

loga x = loga 7+loga 3 -loga 11.

loga x = loga (7*3) - loga 11.

loga x = loga 21 - log 11

loga x = loga (21/11).

Now we take the anti log with respect to base a.

So x = 21/7.

**Therefore x = 21/11.**

To determine x, we'll apply the product and quotient rule of logarithms:

log(b*c) = logb + logc

loga 7+loga 3 = loga(7*3) = loga 21

The expression will become:

loga x= loga 21 - loga 11

log(b/c) = logb - logc

loga 21 - loga 11 = loga(21/11)

The expression will become:

loga x = loga(21/11)

Since the bases are matching, we'll apply the one to one rule:

**x = 21/11**