3-lg(X^2+1)=lg(1/(x+1))+log5 5first 5 is little number, I guess that log5 5 equals 1.

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neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

Posted on

We rewrite the equation as below:

3 = lg(x^2+1)+lg(1/(x+1))+ log(5) 5.

We know log(a) a = 1. So log(5) 5 = 1.

=> 3-1 = lg(x^2+1)+lg(1/(x+1))

=> 2 = lg(x^2+1)/(x+1), as lga+lgb = lgab.

We take antilog:

100 = (x^2+1)/(x+1).

Multiply by (x+1)

=> 100(x+1) = x^2+1

=> x^2-100x+1-100 =0

=> x^2-100x-99 = 0

x = {100+or-sqrt(100^2+4*99)}/2

x = 50 +or-sqrt(50^2+99)

x= 50+sqrt(2599)

 

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

Yes, it is true, log5 5 = 1.

We'llr e-write the given equation:

3 - lg(x^2 + 1) = lg [1/(x+1)] + 1

We'll create matching bases and we'll start by writting 1 = lg 10.

3 - lg(x^2 + 1) = lg [1/(x+1)] + lg 10

We can write 3 = 3*1 = 3*lg 10

We'll use power property:

3*lg 10 = lg 10^3

The equation will become:

lg 10^3 - lg(x^2 + 1) = lg [1/(x+1)] + lg 10

We'll use the quotient property of logarithms to the left side:

lg 1000/(x^2 + 1) = lg [1/(x+1)] + lg 10

We'll use the product property of logarithms to the right side:

lg 1000/(x^2 + 1) = lg [10/(x+1)]

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

1000/(x^2 + 1) =  10/(x+1)

We'll cross multiply:

10(x^2 + 1) = 1000(x+1)

We'll divide by 10:

x^2 + 1 = 100x + 100

We'll subtract 100x + 100 both sides:

x^2 - 100x - 99 = 0

x1 = [100+sqrt(100396)]/2

x1 = 50 + sqrt25099

x2 = 50 - sqrt25099

Only the 1st value of x respects the constraints of existence of logarithms, therefore the equation will have only one solution x = 50 + sqrt25099.

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