# Log 5 (5x+25) - 2 = 0

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log 5 (5x+25) - 2 = 0

Move 2 to the right sides by adding 2:

==> log 5 (5x + 25) = 2

==> 5x + 25 = 5^2

==> 5x + 25 = 25

reduce similar:

==> 5x = 0

==> x= 0

Let us check the answer:

log 5 (5*0 + 25) -2 = 0

log 5 (25) -2 =0

log 5 (5^2) - 2= 0

2log 5 (5) -2 = 0

2 - 2 = 0

0=0

Then the answer is x= 0

To solve log 5 (5x+25) - 2 = 0

Solution:

This is logatithmatice equation of log base 5.

Therefore 2 = log 5 (5^2). log a-log b = log(a/b)

So the equation given becomes:

log 5 (5x+25) - log5 (5^2) = 0

log5 (5x+25) = log5 (5^2). Take antilog to base 5.

5x+25 = 5^2 = 25

5x = 0

x = 0.

Verification:

LHS:

log 5 (5x+25) -2

=log5 (5*0+25) -2

= log5 (25) -2

log5 (5^2) - 2

= 2-2

= 0 = RHS.

For starting, we'll have to impose constraints of existance of logarithm function.

5x+25>0

We'll add -25 both sides:

5x>-25

We'll divide by 5:

x>-5

So, for the logarithms to exist, the values of x have to be in the interval (-5, +inf.)

We'll shift the free term to the right side:

log 5 (5x+25) = 2

We'll create matching bases to the right side.

log 5 (5x+25) = log 5 (5^2)

Now, we'll use the one to one property:

5x+25 = 25

We'll eliminate like terms:

5x = 25-25

5x = 0

We'll divide by 5:

**x = 0 > -5**

The solution is admissible because the value belongs to the interval (-5,+inf.)