Question

Log16 (x) +log4 (x) +log2 (x) =7

Answer 1

Log₁₆X + log₄X + log₂X = 7
ОДЗ: x>0

$\frac{ log_{2}X }{4} + \frac{ log_{2}X }{2} + log_{2}X = 7$

log₂X + 2log₂X + 4log₂X = 28
log₂X  + log₂X² + log₂X⁴ = 28
log₂X⁷ = 28
x⁷ = 2²⁸
x = 2⁴
x= 16
Ответ: 16

Answer 2

To solve the equation log₁₆(x) + log₄(x) + log₂(x) = 7, we can use logarithmic properties to simplify the equation.

Since all three logarithms have different bases, we can use the change of base formula to express them in terms of a common base. Let's choose base 2 for simplicity:

log₁₆(x) = log₂(x) / log₂(16) = log₂(x) / 4 log₄(x) = log₂(x) / log₂(4) = log₂(x) / 2

Now, we can rewrite the equation as:

log₂(x) / 4 + log₂(x) / 2 + log₂(x) = 7

To combine the logarithms, we need a common denominator, which is 4:

(log₂(x) + 2log₂(x) + 4log₂(x)) / 4 = 7

(7log₂(x)) / 4 = 7

Now, we can solve for log₂(x):

7log₂(x) = 4 * 7

7log₂(x) = 28

Dividing both sides by 7:

log₂(x) = 4

To convert this equation to exponential form, we have:

2^4 = x

x = 16

Therefore, the solution to the equation log₁₆(x) + log₄(x) + log₂(x) = 7 is x = 16.