How to solve for x?

Question

log x = log 3 + 2 log x

The answer is x =1/3 but i don't know how to get it. Can anyone help?

Answer 1

log x = log 3 x ²
3 x ² = x
3 x ² - x = 1
x (3 x - 1) = 0
x = 0 , x = 1 / 3
x = 1 / 3 is acceptable value. ( x ≠ 0 )

Answer 2

You need a calculator
Log
x = Log 3 + 2 Log x

Answer 3

x = 1/3

Answer 4

There's a logarithmic law that states that a * log b = log b^a:
log x = log 3 + log x^2

There's a logarithmic law that states that log a + log b = log ab:
log (x) = log (3x^2)

In order for those log equations to be true, the values in the parentheses must be equal.
x = 3x^2

Divide by x.
1 = 3x

Divide by 3.
x = 1/3

Answer 5

Ok, let's see,

2logx = log(x^2)

log 3 + log(x^2)

recall log(ab) = log(a) + log(b)

log(3x^2) = log x

raising both sides to the 10th power,

3x^2 = x
x = 1/3

Good luck!

Answer 6

log(x) = log(3) + 2log(x)

Subtract 2log(x) from both sides:
log(x) - 2log(x) = log(3)

Exponent rule for logs:
log(x) - log(x^2) = log(3)

Division/subtraction rule for logs:
log(x / x^2) = log(3)
log(1 / x) = log(3)

Which implies that:
1 / x = 3
Multiply both sides by x:
1 = 3x
Divide both sides by 3:
x = 1/3

Answer 7

note log=In
In x=In 3+2 In x
In x=In 3 +In(x^2)
In x=In(3x^2)
3x^2=x
x(3x-1)=0
x=1/3 as x cannot equal 0 as In 0 is undefined

Answer 8

log x = log 3x^2 (log ab= loga +log b and log x^k=k log x)
x=3x^2
x-3x^2=0
x(1-3x)=0
1-3x=0
1=3x
x=1/3