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.