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

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