Question

3logx 1/36+log36 1/x=4

Answer 1

logx 1/36=-logx 36

log36 1/x=-log36 x

logx 36=1/logx 36

уравнение получит такой вид

(-3/log36 x)-log36 x=4   log36 x=y

-3/y-y=4

-3-y^2=4y

y^2+4y+3=0

D=4

y1=-2+2=0

y2=-2-2=-4

 

log36 x=1          log36 x=-4

x=36^(0)          x=36^(-4)

x=1              x=1/36^4

Ответ: х1=1   х2=1/(36)^4=1/1679516

Answer 2

We can use the laws of logarithms to simplify the left-hand side of the equation:

3logx(1/36) + log36(1/x) = 4

Using the property loga(b^c) = cloga(b), we can simplify the left-hand side further:

3logx(1/6^2) + log6^2(1/x) = 4

Using the property loga(b/c) = loga(b) - loga(c), we can simplify again:

3[logx(1) - 2logx(6)] + 2[logx(1) - logx(x)] = 4

Simplifying the logarithms of 1 and x:

3[0 - 2logx(6)] + 2[0 - logx(x)] = 4

Simplifying further:

-6logx(6) - 2logx(x) = 4

Using the property loga(b^c) = cloga(b), we can simplify the logarithms:

logx(6^-6) - logx(x^2) = 2

Using the property loga(b/c) = loga(b) - loga(c), we can simplify further:

logx(1/6^6) - logx(x^2) = 2

Using the property loga(b) - loga(c) = loga(b/c), we can simplify again:

logx((1/6^6)/x^2) = 2

Using the property loga(b) = c is equivalent to a^c = b, we can rewrite the equation:

x^2/(1/6^6) = 10^2

Simplifying:

x^2 = 6^12

Taking the square root of both sides:

x = ±6^6

However, we need to check if these solutions are valid since we took the logarithm of x in the original equation. The solution x = -6^6 is not valid since the logarithm of a negative number is undefined. Therefore, the only valid solution is x = 6^6.

Therefore, the solution to the equation is x = 46656.

Answer 3

To solve the equation, we can use the logarithmic rules to simplify it as follows:

3log(x) (1/36) + log(36) (1/x) = 4

log(x^3) (1/36) + log(36) (1/x) = 4

log(x^3/36) + log(36/x) = 4

log(x^3/36 * 36/x) = 4

log(x^3) = 4

x^3 = 10^4

x = 100

Therefore, the solution to the equation is x = 100.