Mister Exam
Other calculators
-
How to use it?
- Equation:
-
Equation 81*x^3+18*x^2+x=0
-
Equation z^3-1=0
-
Equation x*8=25+15
-
Equation (x-1)^4-2*(x-1)^2-3=0
- Express {x} in terms of y where:
- -14*x+10*y=19
- -10*x+15*y=10
- 12*x-3*y=-2
- -11*x-7*y=-2
- Sum of series:
-
16
- Derivative of:
-
16
- Integral of d{x}:
-
16
-
Identical expressions
- four ^(three +x)= sixteen
- 4 to the power of (3 plus x) equally 16
- four to the power of (three plus x) equally sixteen
- 4(3+x)=16
- 43+x=16
- 4^3+x=16
-
Similar expressions
- 4^(3-x)=16
4^(3+x)=16 equation
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
Given the equation:
$$4^{x + 3} = 16$$
or
$$4^{x + 3} - 16 = 0$$
or
$$64 \cdot 4^{x} = 16$$
or
$$4^{x} = \frac{1}{4}$$
- this is the simplest exponential equation
Do replacement
$$v = 4^{x}$$
we get
$$v - \frac{1}{4} = 0$$
or
$$v - \frac{1}{4} = 0$$
Move free summands (without v)
from left part to right part, we given:
$$v = \frac{1}{4}$$
We get the answer: v = 1/4
do backward replacement
$$4^{x} = v$$
or
$$x = \frac{\log{\left(v \right)}}{\log{\left(4 \right)}}$$
The final answer
$$x_{1} = \frac{\log{\left(\frac{1}{4} \right)}}{\log{\left(4 \right)}} = -1$$
$$4^{x + 3} = 16$$
or
$$4^{x + 3} - 16 = 0$$
or
$$64 \cdot 4^{x} = 16$$
or
$$4^{x} = \frac{1}{4}$$
- this is the simplest exponential equation
Do replacement
$$v = 4^{x}$$
we get
$$v - \frac{1}{4} = 0$$
or
$$v - \frac{1}{4} = 0$$
Move free summands (without v)
from left part to right part, we given:
$$v = \frac{1}{4}$$
We get the answer: v = 1/4
do backward replacement
$$4^{x} = v$$
or
$$x = \frac{\log{\left(v \right)}}{\log{\left(4 \right)}}$$
The final answer
$$x_{1} = \frac{\log{\left(\frac{1}{4} \right)}}{\log{\left(4 \right)}} = -1$$
Sum and product of roots
[src]
sum
pi*I
-1 + -1 + ------
log(2)
$$-1 + \left(-1 + \frac{i \pi}{\log{\left(2 \right)}}\right)$$
=
pi*I
-2 + ------
log(2)
$$-2 + \frac{i \pi}{\log{\left(2 \right)}}$$
product
/ pi*I \ -|-1 + ------| \ log(2)/
$$- (-1 + \frac{i \pi}{\log{\left(2 \right)}})$$
=
pi*I
1 - ------
log(2)
$$1 - \frac{i \pi}{\log{\left(2 \right)}}$$
1 - pi*i/log(2)
Rapid solution
[src]
x1 = -1
$$x_{1} = -1$$
pi*I
x2 = -1 + ------
log(2)
$$x_{2} = -1 + \frac{i \pi}{\log{\left(2 \right)}}$$
x2 = -1 + i*pi/log(2)