If Log2 x - 5 Log x + 6 = 0, then what would the value / values of x be?

Question

plz solve with explanation.

Answer 1

The notation is ambiguous here.

If log2 x means (log x)^2, the answer can be found by making the substitution y = log x to give
y^2 - 5y + 6 = 0, which gives the solutions
y = 3
y = 2

Assuming that the logs are to the base 10, substituting back gives

log x = 3 -> x = 10^3 -> x = 1000
log x = 2 -> x = 10^2 -> x = 100

If log2 x means log to the base 2 of x, the equation is still solvable but the answer is messier:

log2 x - 5 log x + 6 = 0
(log x) / (log 2) - 5 log x = -6
(log x) (1/log 2 - 5) = -6
log x = -6 / (1/log 2 - 5)
log x = (6 log 2) / (5 log 2 - 1)
log x = log 64 / (log 32 - 1)
x = 64^(1 / (log 32 - 1))

If the logarithm is to the base 10,

x ≈ 3763

Answer 2

Assuming you mean log-base-2:

log2(x) - 5 log(x) + 6 = 0

Use the base conversion formula to write log2(x) in terms of log(x):

log(x)/log(2) - 5 log(x) = -6

Use distributive property to factor out log(x) and then divide to isolate it:

(1/log(2) - 5) log(x) = -6
log(x) = -6 / (1/log(2) - 5)

Everything on the right is a constant, so we can just solve for log(x):

log(x) = -6 / (3.3219 - 5)
log(x) = 3.575521

And then the inverse of common log (10^x) can be used to tell us what x is:

x = 10^(3.575521)
x = 3,762.9818

Answer 3

is this Log(2 x) or Log x base 2