# log100=2(logx+log5) What is x?

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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.

Take antilog:

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:

log100=2(logx+log5)

log100=2logx +2log5

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.**