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What is the true solution to the logarithmic equation below? log4[log4(2x]=1
x=2
x=8
x=65
x=128

Respuesta :

1. 
write 1 as [tex]log_{4}4[/tex] so that we compare same base logarithms:

[tex] log_{4}[{ log_{4}}2x]= log_{4}4[/tex]

2. Neglect the equal bases and write the arguments:
 
So [tex]{ log_{4}}2x=4[/tex]

repeat step 1, that is write 4 as a logarithm with base 4: 

 [tex]4=4*1=4* log_{4}4= log_{4} 4^{4} [/tex]

3. [tex]{ log_{4}}2x=log_{4} 4^{4}[/tex]

      [tex]2x=4^{4}[/tex]
       
  [tex]x= \frac{4^{4}}{2}= 4^{3}*2=64*2=128[/tex]
aksnkj

The true solution to the logarithmic equation log4[log4(2x] = 1 will be x = 128.

To solve the equation we will be using Logarithm.

Given :

log4[log4(2x] = 1

How to solve the logarithmic equation?

Logarithmic equations are the equations containing log expression.

We will first write the equation [tex]log_{4}4[/tex] then we compare the same base logarithms,

[tex]\rm log_{4}[log_{4}2x] = log_{4}4[/tex]

Now, we neglect the equal bases & then we write the arguments.

[tex]\rm log _{4}2x=4[/tex]

Now, we repeat first line then we write 4 as a logarithm with base 4:

[tex]\rm 4= 4\times 1 =4 \times log_{4}4=log_{4}4^4[/tex]

On solving further we get,

[tex]\rm log_{4}2x=log_{4}4^4\\\\2x=4^4\\\\x=\dfrac{4^4}{2}=4^3\times2=64\times2=128[/tex]

Therefore, The true solution to the logarithmic equation log4[log4(2x] = 1 will be x = 128.

Learn more about Logarithm here:

https://brainly.com/question/4102150

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