CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

Solve the following systems of equations.
log3(log2x)+log1/3(log1/2y)=1,xy24,

Open in App
Solution

Solution: Given,

=> log₃(log₂x) + log¹/₃ (ˡᵒᵍ¹/₂ ʸ) = 1

=> log₃(log₂x) + -1log₃(-1log₂y) = 1 [∴ logₐ⁻ⁿ b = -nlogb ]

=> log₃(log₂x) + log₃(log₂ y⁻¹)⁻¹ = 1 [∴ nloga = logaⁿ]

=> log₃(log₂x)•(log₂y⁻¹)⁻¹ = 1 [∴ loga + logb = logab ]

=> log₃(log₂x)•(log₂y⁻¹)⁻¹ = log₃(3) [∴ logₐ a = 1]

=> (log₂x)•(log₂y⁻¹)⁻¹ = 3 [ ∴ loga = logb , a = b]

=> (log₂x)/(log₂y⁻¹) = 3

=> log₂(x) = log₂(y⁻³)

=> x = y⁻³

=> x = 1/y³

• But, here the second equation →( xʸ² 4 ) is incomplete or having error .

• Therefore, we cannot find the value of X independent of Y for the solution of equation.

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon