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Identical expressions
- log two (2x- one)=2
- logarithm of 2(2x minus 1) equally 2
- logarithm of two (2x minus one) equally 2
- log22x-1=2
-
Similar expressions
- log2(2x+1)=2
- log(2)(x²-1)=log(2)(2x-1)
- log^(2)(2x-1)=3
log2(2x-1)=2 equation
The teacher will be very surprised to see your correct solution 😉
The solution
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[src]
log(2*x - 1) ------------ = 2 log(2)
$$\frac{\log{\left(2 x - 1 \right)}}{\log{\left(2 \right)}} = 2$$
Detail solution
Given the equation
$$\frac{\log{\left(2 x - 1 \right)}}{\log{\left(2 \right)}} = 2$$
$$\frac{\log{\left(2 x - 1 \right)}}{\log{\left(2 \right)}} = 2$$
Let's divide both parts of the equation by the multiplier of log =1/log(2)
$$\log{\left(2 x - 1 \right)} = 2 \log{\left(2 \right)}$$
This equation is of the form:
By definition log
then
$$2 x - 1 = e^{\frac{2}{\frac{1}{\log{\left(2 \right)}}}}$$
simplify
$$2 x - 1 = 4$$
$$2 x = 5$$
$$x = \frac{5}{2}$$
$$\frac{\log{\left(2 x - 1 \right)}}{\log{\left(2 \right)}} = 2$$
$$\frac{\log{\left(2 x - 1 \right)}}{\log{\left(2 \right)}} = 2$$
Let's divide both parts of the equation by the multiplier of log =1/log(2)
$$\log{\left(2 x - 1 \right)} = 2 \log{\left(2 \right)}$$
This equation is of the form:
log(v)=p
By definition log
v=e^p
then
$$2 x - 1 = e^{\frac{2}{\frac{1}{\log{\left(2 \right)}}}}$$
simplify
$$2 x - 1 = 4$$
$$2 x = 5$$
$$x = \frac{5}{2}$$
Sum and product of roots
[src]
sum
0 + 5/2
$$0 + \frac{5}{2}$$
=
5/2
$$\frac{5}{2}$$
product
1*5/2
$$1 \cdot \frac{5}{2}$$
=
5/2
$$\frac{5}{2}$$
5/2
The graph