# log 2x + 2log 5 = 2

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log 2x + 2 log 5 = 2

We knwo that:

alog b = log b^a

==? log 2x + log 5^2 = 2

==> log 2x + log 25 = 2

Also, we know that:

log a + log b = log ab

==> log 2x*25 = 2

==> log 50x = 2

==> 50x = 10^2

==> 50x = 100

==> x= 100/50 = 2

==> x = 2

Before solving the equation, we'll have to impose constraints of existence of logarithms.

2x>0

x>0

Now, we could apply the power property of logarithms:

2log 5 = log 5^2 = log 25

We'll re-write the equation:

log 2x + log 25 = 2

Now, we'll apply the product property of logarithms:

log 2x + log 25 = log 2*25*x = log 50x

log 50x = 2 => 50x = 10^2 => 50x = 100

We'll divide by 50 both sides:

x = 2 >0

Because the solution is in the interval (0, +inf.), the solution is admissible.

log2x+2log5 = 2.

We uase the property of logarithms:

m*loga = loga^m ; log a - log b = log(a/b) and log 10^m = m log 10 = m*1 = m. Or m = log10M

Given equation is:

log2x + 2log5 = 2.

log2x +log5^2 = log 10^2. Subtract log5^2.

log2x = log10^2 - log5^2

log2x = log (10^2/5^2). Take antilogarithms.

2x = 10^2/5^2 = 100/25 = 4

2x/2 = 4/2

x = 2.

Verification:

LHS=log2x+2log5

= log2*2 +2log5

= log4+log25

= log(4*25)

=log100

=log(10^2) = 2 by definition of logarithm.

= RHS.