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log 100 = 2(logx +log5). To solve for x.
WE use the property of logarithms to solve this:
loga+logb = logab ,
loga - logb = loga/b and
m loga = loga^m.
Therefore in the given equation, 2 (logx+log5) = 2log5x
2(logx+5) = log(5x)^2
2(logx+5) = log(25x^2)
Therefore the given equation becomes:
log100 = log25x^2.
100 = 25x^2
100/25 = x^2
4 = x^2
x = 2. or x = -2. But -2 is invalid as log-2 is undefined in real numbers
Therefore x = 2.
We'll remove the brackets from the right side:
First, we'll use the power property of logarithms, for the terms of the expression:
2 log 5 = log 5^2
2log x = log x^2
We'll re-write the expression:
log100 = log x^2 + log 5^2
Since the bases are matching, we'll use the product property of logarithms:
log a + log b = log a*b
We'll put a = x^2 and b = 5^2
log x^2 + log 5^2 = log x^2*5^2
We'll write the equation:
log 100 = log x^2*5^2
Since the bases are matching, we'll apply one to one property:
100 = x^2*5^2
We'll use symmetric property:
25x^2 = 100
We'll divide by 25;
x^2 = 4
x1 = -2
x2 = 2
Since the solution of the equtaion must be positive, the first solution x1 = -2, will be rejected.
The equation will have just a solution, x = 2.
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