What is the true solution to the logarithmic equation?
log₂[log₂ (√4x)] = 1
1 answer:
Note that
•
is defined as long as ![x\ge0](https://tex.z-dn.net/?f=x%5Cge0)
•
is defined as long as ![x>0](https://tex.z-dn.net/?f=x%3E0)
•
is defined as long as
![\log_2(x) > 0 \implies 2^{\log_2(x)} > 2^0 \implies x > 1](https://tex.z-dn.net/?f=%5Clog_2%28x%29%20%3E%200%20%5Cimplies%202%5E%7B%5Clog_2%28x%29%7D%20%3E%202%5E0%20%5Cimplies%20x%20%3E%201)
so that overall, any solution to this equation must be larger than 1.
In the equation itself, introduce powers of 2 and squares to eliminate the logarithms and square root. Solve for
.
![\log_2\left(\log_2\left(\sqrt{4x}\right)\right) = 1](https://tex.z-dn.net/?f=%5Clog_2%5Cleft%28%5Clog_2%5Cleft%28%5Csqrt%7B4x%7D%5Cright%29%5Cright%29%20%3D%201)
![2^{\log_2\left(\log_2\left(\sqrt{4x}\right)\right)} = 2^1](https://tex.z-dn.net/?f=2%5E%7B%5Clog_2%5Cleft%28%5Clog_2%5Cleft%28%5Csqrt%7B4x%7D%5Cright%29%5Cright%29%7D%20%3D%202%5E1)
![\log_2\left(\sqrt{4x}\right) = 2](https://tex.z-dn.net/?f=%5Clog_2%5Cleft%28%5Csqrt%7B4x%7D%5Cright%29%20%3D%202)
![2^{\log_2\left(\sqrt{4x}\right)} = 2^2](https://tex.z-dn.net/?f=2%5E%7B%5Clog_2%5Cleft%28%5Csqrt%7B4x%7D%5Cright%29%7D%20%3D%202%5E2)
![\sqrt{4x} = 4](https://tex.z-dn.net/?f=%5Csqrt%7B4x%7D%20%3D%204)
![\left(\sqrt{4x}\right)^2 = 4^2](https://tex.z-dn.net/?f=%5Cleft%28%5Csqrt%7B4x%7D%5Cright%29%5E2%20%3D%204%5E2)
![4x = 16](https://tex.z-dn.net/?f=4x%20%3D%2016)
![\boxed{x = 4}](https://tex.z-dn.net/?f=%5Cboxed%7Bx%20%3D%204%7D)
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