help with this math problem?

Question

2 logbase3x+ logbase314=12

that is log base 3 14. not logbase314

Answer 1

log_3 x^2 + log_3 14 = 12

log_3 (14x^2) = 12

3^12 = 14x^2

x = +/- [(3^6)/sqrt(14)]

Answer 2

2 logbase3x+ logbase314=12

Because the two logs have same base, we can add them. But remember, when you add them, the arguments inside the log acutally multiply.

2 logbase3(x*14)=12
logbase3(x*14)=6

14x=6^3
x=108/7

Answer 3

2 log x + log 14 = 12
since 2 log x = log (x^2)
log(x)^2 + log 14 = 12
since log a + log b = log (ab)
log (x^2)(14) = 12
this log is to the base 3
3^12 = 14 x^2
x^2 = 3^12/14
x = sqrt[(3^12)/14]
= (3^6)sqrt(14) = 729/3.741
= 194.86

Answer 4

I think you are saying 2log(x) + log(14) = 12, where the logs are base 3, right?

2log(x) + log(14) =
log(x^2) + log(14) =
log(14x^2) = 12
Raise both sides to the power 3:
3^log(14x^2) = 3^12
14x^2 = 3^12
x^2 = (3^12)/14
x = (3^6)/SQRT(14)

Answer 5

Let's use L3 to mean log base 3

If I understand the question you have:
2L3(x) + L3(14)=12

2L3(x)=L3(x²), and L3(a)+L3(b) = L3(ab) so the left side looks like
L3(14x²)=12

Now, let's look at what L3(a)=n really means.
It means that 3ⁿ=a

With that in mind,
3¹²=14x²
14x²-3¹²=0... and you have the difference of two not-so-perfect squares.
(x√(14)+3⁶)(x√(14)-3⁶)

Solve for x... Have to tell you, it's not pretty, but it can be done. You'll either have to use a calculator, or a spreadsheet.

Answer 6

Rewrite
log3[x^2] = 12 - log3[14]
Elevate to 3
x^2 = 3^{12 - log3[14]} = (3^12){14^(-1)} = (3^12)/14
x = +-(3^6)/sqr[14]

Answer 7

2log(3)x + log(3)14 = 12
log(3)x² + log(3)14=
log(3) (14x²)=12
3^12=14x²
x²=3^12/14
x=± 3^6/√14 ≈ ±194.8

Answer 8

2 logbase3 x+ logbase3 14=12
log base3 (14x^2) =12
14x^2 = 3^12
x^2 = 3^12/14
x = 194.83