# log_10(x^3-1)-log_10(x^2+x+1)=-2

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### 4 Answers

log 10 (x^3 -1) - log 10 (x^2 + x+1) = -2

We know that:

log a - log b = log a/b

==> log 10 (x^3 -1)/(x^2 + x +1) = -2

But we know that:

x^3 -1 = (x-1)(x^2 + x+1)

==> log (x-1)(x^2 + x+1)/(x^2 + x+1) = -2

Now reduce similar :

==> log 10 (x-1) = -2

==> x-1 = 10^-2

==> x-1 = 1/10^2

==> x-1= 1/100

==> x-1 = 0.01

==> x = 1+0.01

**==> x= 1.01**

Here we use the relation that log A - log B = log( A/B).

( log in the solution refers to logarithm to the base 10. )

Therefore: log (x^3-1)- log ( x^2+x+1)= -2

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

Now take the antilog of both the sides. For -2 the antilog is 10^-2=1/100.

=> (x^3-1)/(x^2+x+1)=1/100

Now x^3-1= (x-1)(x^2+x+1) can be used to simplify the expression.

=> (x-1)(x^2+x+1)= (1/100)*(x^2+x+1)

Cancelling (x^2+x+1)

=> x-1 =(1/100)

=> x= (1/100)+1= 1.01

**Therefore x= 1.01**

Let's solve the problem trying another approach. First, we'll write the value 2 from the right side as it follows:

-2 = -2*1

We'll substitute the factor 1 by the equivalent log_10 (10).

-2 = -2*log_10 (10)

We'll use the power rule of logarithms:

-2 = log_10 (10)^-2

Now, we'll re-write the equation substituting -2 by the log_10 (10)^-2:

log_10(x^3-1) - log_10(x^2+x+1) = - log_10 (10)^-2

We'll subtract log_10(x^2+x+1) both sides:

log_10(x^3-1) = log_10(x^2+x+1) - log_10 (10)^-2

We'll spply the quotient rule of logarithms, to the righht side (the difference of logarithms is the logarithm of quotient):

log_10(x^3-1) = log_10 [(x^2+x+1)/(10)^-2]

Now, because the bases are matching, we'll use the one to one property of logarithms:

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

But 10^-2 = 1/10^2 = 1/100

(x^3-1) = 100[(x^2+x+1)]

We'll re-write the difference of cubes from the left side:

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

Now, we'll re-write the equation:

(x-1)[(x^2+x+1)] = 100[(x^2+x+1)]

Now, we'll divide by [(x^2+x+1)] both sides:

x-1 = 100

We'll add 1 both sides:

x = 100+1

**x = 101**

To solve log 10 (x^3-1) - log 10(x^2+x+1) = -2, wher 10 is treated as the base of logarithm.

We know that -2 = log 10^-2 , by definition as log 10^x = x.

So the given equation becomes:

log 10 { (x^3-1) /(x^2+x+1) } = log10^-2 , as loga-log b = loga/b.

Take the antilog.

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

(x-1)(x^2+x+1)/(x^2+x+1) = 10^-2, as a^3-b^3 = (a-b)(a^2+ab+b^2) is an identity.

Therefore x-1 = 10^-2 = 1/100.

x= 1+1/100 = 1.01.

Hope this helps.