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3 years ago

What is the true solution to the logarithmic equation?log2 [log2(sqrt4x)]=1

Mathematics

2 answers:

erastova [34]3 years ago

8 0

Answer:4

Step-by-step explanation:

log₂[log₂(√4x)] = 1

log₂2 =1

So we replace our 1 with log₂2

log₂[log₂(√4x)] = log₂2

log₂ on bothside will cancel each other.

We will be left with;

[log₂(√4x)] = 2

log = power of exponential

√4x = 2²

√4x = 4

Square bothside

(√4x)² = 4²

4X = 16

Divide bothside by 4

4x/4 = 16/4

x = 4

Deffense [45]3 years ago

6 0

Answer:

x=4

Step-by-step explanation:

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Step-by-step explanation:

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2 years ago

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Step-by-step explanation:

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3 years ago

What is the true solution to the logarithmic equation?log2 [log2(sqrt4x)]=1​

What is the true solution to the logarithmic equation?log2 [log2(sqrt4x)]=1