# Solution of logarithmic equationslog(4x+16)=log100-2log2+log16

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We have to solve log(4x+16) = log100 - 2log2 + log16

Use the relations log a + log b = log a*b, log a - log b = log(a/b) and a*log b = log b^a

log(4x+16) = log100 - 2log2 + log16

=> log(4x+16) = log100 - log 2^2 + log 16

=> log(4x+16) = log(100*16/4)

=> log(4x+16) = log(400)

Equate 4x + 16 = 400

=> x + 4 = 100

=> x = 96

The required value of x is x = 96

We'll shift log 100 to the left side:

log(4x+16) - log100 = log16 - 2*log2

We'll apply power rule of logarithms:

log(4x+16) - log100 = log16 - log 2^2

We'll apply quotient rule, both sides of the equation:

log [(4x+16)/100] = log(16/4)

Since the bases are matching, we'll apply one to one property:

[(4x+16)/100] = 16/4

[(4x+16)/100] = 4

We'll multiply by 100 both sides:

4x + 16 = 4*100

We'll divide by 4:

x + 4 = 100

We'll subtract 4:

x = 100 - 4

x = 96

The constraint of existence of logarithm is 4x+16>0

4x>-16

x>-4

The interval of admissible values for x is (-4 ; +infinite).

**Since the value of x belongs to the interval of admissible values, we'll accept as solution of the equation x = 96.**