log(2x) = log(x+5) =2

We know that logx + logy= log(xy)

==> log(2x) + log(x+5) = log(2x(x+5)

==> log(2x^2 +10x) = 2

But log100 = 2

==> log(2x^2 + 10x) = log 100

==> 2x^2 + 10x = 100

Divide by 2:

==> x^2 + 5x = 50

Subtract 50...

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log(2x) = log(x+5) =2

We know that logx + logy= log(xy)

==> log(2x) + log(x+5) = log(2x(x+5)

==> log(2x^2 +10x) = 2

But log100 = 2

==> log(2x^2 + 10x) = log 100

==> 2x^2 + 10x = 100

Divide by 2:

==> x^2 + 5x = 50

Subtract 50 from both sides:

==> x^2 +5x -50=0

Factorize:

==> (x+10)(x-5) =0

==> x1= 5

==> x2= -10 ( this is impossible because the function is not defined)

**Then the solution is x= 5**