What is x if loga x= loga 7+loga 3-loga 11?
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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
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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
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