Mister Exam
Other calculators
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How to use it?
- Equation:
-
Equation x^2+x=12
-
Equation 3x-2(3x-2)=19
-
Equation x^4=(x-56)^2
-
Equation (-2x+1)*(-2x-7)=0
- Express {x} in terms of y where:
- -7*x-13*y=-12
- 8*x+1*y=4
- 20*x-16*y=7
- -3*x+19*y=16
- Derivative of:
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e^x
- Integral of d{x}:
-
e^x
- Graphing y =:
- e^x
-
Identical expressions
- e^x=t²- one
- e to the power of x equally t² minus 1
- e to the power of x equally t² minus one
- ex=t²-1
-
Similar expressions
- e^x=t²+1
e^x=t²-1 equation
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
Given the equation:
$$e^{x} = t^{2} - 1$$
or
$$\left(1 - t^{2}\right) + e^{x} = 0$$
Do replacement
$$v = e^{x}$$
we get
$$- t^{2} + v + 1 = 0$$
or
$$- t^{2} + v + 1 = 0$$
Move free summands (without v)
from left part to right part, we given:
$$- t^{2} + v = -1$$
Divide both parts of the equation by (v - t^2)/v
We get the answer: v = -1 + t^2
do backward replacement
$$e^{x} = v$$
or
$$x = \log{\left(v \right)}$$
The final answer
$$x_{1} = \frac{\log{\left(t^{2} - 1 \right)}}{\log{\left(e \right)}} = \log{\left(t^{2} - 1 \right)}$$
$$e^{x} = t^{2} - 1$$
or
$$\left(1 - t^{2}\right) + e^{x} = 0$$
Do replacement
$$v = e^{x}$$
we get
$$- t^{2} + v + 1 = 0$$
or
$$- t^{2} + v + 1 = 0$$
Move free summands (without v)
from left part to right part, we given:
$$- t^{2} + v = -1$$
Divide both parts of the equation by (v - t^2)/v
v = -1 / ((v - t^2)/v)
We get the answer: v = -1 + t^2
do backward replacement
$$e^{x} = v$$
or
$$x = \log{\left(v \right)}$$
The final answer
$$x_{1} = \frac{\log{\left(t^{2} - 1 \right)}}{\log{\left(e \right)}} = \log{\left(t^{2} - 1 \right)}$$
The graph
Sum and product of roots
[src]
sum
/ 2\ 0 + log\-1 + t /
$$\log{\left(t^{2} - 1 \right)} + 0$$
=
/ 2\ log\-1 + t /
$$\log{\left(t^{2} - 1 \right)}$$
product
/ 2\ 1*log\-1 + t /
$$1 \log{\left(t^{2} - 1 \right)}$$
=
/ 2\ log\-1 + t /
$$\log{\left(t^{2} - 1 \right)}$$
log(-1 + t^2)