# If log x^3- log 10x = log 10^5 find x

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log x^3- log 10x = log 10^5

We will use logarithm properties to solve for x.

We know that log a - log b = log a/b

==> log x^3 - log 10x = log x^3/10x = log x^2/10

==> log x^2/10 = log 10^5

Now we have two equal logarithm with equal bases. Then, we conclude that the logs are equal.

==> x^2 /10 = 10^5

Now let us multiply both sides by 10.

==> x^2 = 10^5 * 10

We know that x^a * x^b = x^(a+ b)

==> x^2 = 10^(5+ 1)

==> x^2 = 10^6

Now we will re-write 10^6.

We know that 10^6 = 10^(2*3) = (10^3)^2.

==> x^2 = ( 10^3) ^2

Now we will take the root of both sides.

==> x = 10^3= 1000

**==> Then, the answer is x = 1000.**

We'll impose costraints of existence of logarithms:

x^3>0 => x>0

10x>0 => x>0

We'll solve the equation adding log 10x both sides:

log x^3 = log 10x + log 10^5

We'll apply the product rule of logarithms:

log x^3 = log x*10^6

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

x^3 = x*10^6

We'll subtract 10^6x both sides:

x^3 - 10^6*x = 0

We'll factorize by x:

x(x^2 - 10^6) = 0

We'll set each factor as zero:

x = 0

We'll reject this answer since x>0.

x^2 - 10^6 = 0

We'll re-write the difference of squares:

(x - 1000)(x + 1000) = 0

We'll set each factor as zero;

x - 1000 = 0

x = 1000

x + 1000 = 0

x = -1000

Since this answer is negative, we'll reject it, too.

**The only valid solution of the equation is x = 1000.**

If log x^3- log 10x = log 10^5 find x.

We know lga - logb = log(a/b).

So LHS = log x^3- log 10x = log(x^3/x)

log x^3- log 10x = log(x^2) .

Therefore log x^3- log 10x = log(x^3/10x)

log(x^3/10x) = log(x^2/10) = RHS = log10^5

Therefore log(x^3/10x) = log10^5.

WE take anti logarithms:

x^3/10 = 10^5.

Multiply by 10 and we get:

x^3 = 10^5*10 = 10^6, a^m*a^n = a^(m+n)

x^3 = 10^6.

x ^3 = (10^2)^3.

Take cube root.

x = 10^2 = 100.

Therefore x= 100.