log2(2x-1)=2 equation
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The solution
Detail solution
Given the equation
log ( 2 x − 1 ) log ( 2 ) = 2 \frac{\log{\left(2 x - 1 \right)}}{\log{\left(2 \right)}} = 2 log ( 2 ) log ( 2 x − 1 ) = 2 log ( 2 x − 1 ) log ( 2 ) = 2 \frac{\log{\left(2 x - 1 \right)}}{\log{\left(2 \right)}} = 2 log ( 2 ) log ( 2 x − 1 ) = 2 Let's divide both parts of the equation by the multiplier of log =1/log(2)
log ( 2 x − 1 ) = 2 log ( 2 ) \log{\left(2 x - 1 \right)} = 2 \log{\left(2 \right)} log ( 2 x − 1 ) = 2 log ( 2 ) This equation is of the form:
log(v)=p By definition log
v=e^p then
2 x − 1 = e 2 1 log ( 2 ) 2 x - 1 = e^{\frac{2}{\frac{1}{\log{\left(2 \right)}}}} 2 x − 1 = e l o g ( 2 ) 1 2 simplify
2 x − 1 = 4 2 x - 1 = 4 2 x − 1 = 4 2 x = 5 2 x = 5 2 x = 5 x = 5 2 x = \frac{5}{2} x = 2 5
The graph
-10.0 -7.5 -5.0 -2.5 0.0 2.5 5.0 7.5 10.0 12.5 15.0 17.5 -20 20
x 1 = 5 2 x_{1} = \frac{5}{2} x 1 = 2 5
Sum and product of roots
[src]
1 ⋅ 5 2 1 \cdot \frac{5}{2} 1 ⋅ 2 5