Mister Exam
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How to use it?
- Equation:
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Equation 4-6*(x+2)=3-5*x
-
Equation 2-3(7+2x)=11
-
Equation x^3=27
-
Equation -x-4+5(x+3)=5(-1-x)-2
- Express {x} in terms of y where:
- -11*x+15*y=7
- 4*x+2*y=5
- 1*x+10*y=-18
- -17*x+1*y=-11
-
Identical expressions
- log2(four -x)= seven
- logarithm of 2(4 minus x) equally 7
- logarithm of 2(four minus x) equally seven
- log24-x=7
-
Similar expressions
- (2sqrt3(pi*x+3*pi)-tg(pi*x-pi/2))*log2(4-x^2)=0
- log2(4+x)=7
log2(4-x)=7 equation
The teacher will be very surprised to see your correct solution 😉
The solution
You have entered
[src]
log(4 - x) ---------- = 7 log(2)
$$\frac{\log{\left(4 - x \right)}}{\log{\left(2 \right)}} = 7$$
Detail solution
Given the equation
$$\frac{\log{\left(4 - x \right)}}{\log{\left(2 \right)}} = 7$$
$$\frac{\log{\left(4 - x \right)}}{\log{\left(2 \right)}} = 7$$
Let's divide both parts of the equation by the multiplier of log =1/log(2)
$$\log{\left(4 - x \right)} = 7 \log{\left(2 \right)}$$
This equation is of the form:
By definition log
then
$$4 - x = e^{\frac{7}{\frac{1}{\log{\left(2 \right)}}}}$$
simplify
$$4 - x = 128$$
$$- x = 124$$
$$x = -124$$
$$\frac{\log{\left(4 - x \right)}}{\log{\left(2 \right)}} = 7$$
$$\frac{\log{\left(4 - x \right)}}{\log{\left(2 \right)}} = 7$$
Let's divide both parts of the equation by the multiplier of log =1/log(2)
$$\log{\left(4 - x \right)} = 7 \log{\left(2 \right)}$$
This equation is of the form:
log(v)=p
By definition log
v=e^p
then
$$4 - x = e^{\frac{7}{\frac{1}{\log{\left(2 \right)}}}}$$
simplify
$$4 - x = 128$$
$$- x = 124$$
$$x = -124$$
Sum and product of roots
[src]
sum
-124
$$-124$$
=
-124
$$-124$$
product
-124
$$-124$$
=
-124
$$-124$$
-124
The graph