Mister Exam
Other calculators
-
How to use it?
- Equation:
- Equation x-y=1
-
Equation 4-6(x+2)=3-5x
-
Equation 1/(x-3)^2-3/(x-3)-4=0
-
Equation 14/(x-3)=-7/9
- Express {x} in terms of y where:
- 4*x+2*y=5
- 1*x+10*y=-18
- 13*x-15*y=-6
- 4*x-7*y=4
- Integral of d{x}:
-
2^x
- Derivative of:
-
2^x
- Graphing y =:
- 2^x
- Canonical form:
- 32
- Expression:
- 32
- 32
-
Identical expressions
- two ^x= thirty-two
- 2 to the power of x equally 32
- two to the power of x equally thirty minus two
- 2x=32
-
Similar expressions
- 3x+29=4x-3(2x-3)
- 2^x=16^2
- (1/2)^x=3x+5
- (3/2)^x=16/81
- (x^2-3)^2-4=0
- 3(2x-4)=2x-(5x+9)
- 1/(x-3)^2-3/(x-3)-4=0
- 8/4^x-6*2^x+1=0
2^x=32 equation
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
Given the equation:
$$2^{x} = 32$$
or
$$2^{x} - 32 = 0$$
or
$$2^{x} = 32$$
or
$$2^{x} = 32$$
- this is the simplest exponential equation
Do replacement
$$v = 2^{x}$$
we get
$$v - 32 = 0$$
or
$$v - 32 = 0$$
Move free summands (without v)
from left part to right part, we given:
$$v = 32$$
We get the answer: v = 32
do backward replacement
$$2^{x} = v$$
or
$$x = \frac{\log{\left(v \right)}}{\log{\left(2 \right)}}$$
The final answer
$$x_{1} = \frac{\log{\left(32 \right)}}{\log{\left(2 \right)}} = 5$$
$$2^{x} = 32$$
or
$$2^{x} - 32 = 0$$
or
$$2^{x} = 32$$
or
$$2^{x} = 32$$
- this is the simplest exponential equation
Do replacement
$$v = 2^{x}$$
we get
$$v - 32 = 0$$
or
$$v - 32 = 0$$
Move free summands (without v)
from left part to right part, we given:
$$v = 32$$
We get the answer: v = 32
do backward replacement
$$2^{x} = v$$
or
$$x = \frac{\log{\left(v \right)}}{\log{\left(2 \right)}}$$
The final answer
$$x_{1} = \frac{\log{\left(32 \right)}}{\log{\left(2 \right)}} = 5$$
The graph