Using logarithms, it is found that since both sides have different bases, 3x - 1 is not equal to 8.
Using change of base, it is found that 3x - 1 is equals to [tex]\frac{\log{8}}{\log{2}}[/tex]
The equation given is:
[tex]\log{(3x - 1)} = \log_{2}{8}[/tex]
The logarithms at each side of the equality have different bases, hence 3x - 1 is not equal to 8.
Making the conversion of the right side to base 10, we have that:
[tex]\log_{2}{8} = \frac{\log{8}}{\log{2}}[/tex]
Hence:
[tex]\log{(3x - 1)} = \frac{\log{8}}{\log{2}}[/tex]
3x - 1 is equals to [tex]\frac{\log{8}}{\log{2}}[/tex]
A similar problem is given at https://brainly.com/question/24286043
3x-1 is not equal to 8 because the base of the log is different and the value of 3x-1 is equal to 1000.
Given:
The given equation is [tex]log(3x - 1) = log_28[/tex].
In the given equation, the base of logarithm is different on left hand side and right hand side.
The base on left hand side is 10 (general), and that on right hand side is 2.
So, the entity inside the log will not be equal. So,
[tex]3x-1\neq 8[/tex]
Now, change the base of the log and relate the relation as,
[tex]log(3x - 1) = log_28\\log(3x - 1) = \dfrac{log8}{log 2}\\=3\\3x-1=antilog 3=10^3\\3x-1=1000[/tex]
Therefore, 3x-1 is not equal to 8 because the base of the log is different and the value of 3x-1 is equal to 1000.
For more details, refer to the link:
https://brainly.com/question/20838017
