Mister Exam
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How to use it?
- Equation:
-
Equation 2(x+4)(x+2)=x^2+2x
-
Equation 16-3x^2+14x=5x-14
- Equation log7(3-x)=2log7(4)
- Equation x1+x2+x3+x4+x5+x6+x7+x8=0
- Express {x} in terms of y where:
- (x^2+y^2-1)^3+x^2*y^3=0
- -1*x-19*y=-10
- 5*x+5*y=20
- 4*x-16*y=6
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Identical expressions
- log7(three -x)=2log7(four)
- logarithm of 7(3 minus x) equally 2 logarithm of 7(4)
- logarithm of 7(three minus x) equally 2 logarithm of 7(four)
- log73-x=2log74
-
Similar expressions
- log7(3+x)=2log7(4)
log7(3-x)=2log7(4) equation
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The solution
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log(3 - x) log(4) ---------- = 2*------ log(7) log(7)
$$\frac{\log{\left(3 - x \right)}}{\log{\left(7 \right)}} = 2 \frac{\log{\left(4 \right)}}{\log{\left(7 \right)}}$$
Detail solution
Given the equation
$$\frac{\log{\left(3 - x \right)}}{\log{\left(7 \right)}} = 2 \frac{\log{\left(4 \right)}}{\log{\left(7 \right)}}$$
$$\frac{\log{\left(3 - x \right)}}{\log{\left(7 \right)}} = \frac{2 \log{\left(4 \right)}}{\log{\left(7 \right)}}$$
Let's divide both parts of the equation by the multiplier of log =1/log(7)
$$\log{\left(3 - x \right)} = 2 \log{\left(4 \right)}$$
This equation is of the form:
By definition log
then
$$3 - x = e^{\frac{2 \log{\left(4 \right)} \frac{1}{\log{\left(7 \right)}}}{\frac{1}{\log{\left(7 \right)}}}}$$
simplify
$$3 - x = 16$$
$$- x = 13$$
$$x = -13$$
$$\frac{\log{\left(3 - x \right)}}{\log{\left(7 \right)}} = 2 \frac{\log{\left(4 \right)}}{\log{\left(7 \right)}}$$
$$\frac{\log{\left(3 - x \right)}}{\log{\left(7 \right)}} = \frac{2 \log{\left(4 \right)}}{\log{\left(7 \right)}}$$
Let's divide both parts of the equation by the multiplier of log =1/log(7)
$$\log{\left(3 - x \right)} = 2 \log{\left(4 \right)}$$
This equation is of the form:
log(v)=p
By definition log
v=e^p
then
$$3 - x = e^{\frac{2 \log{\left(4 \right)} \frac{1}{\log{\left(7 \right)}}}{\frac{1}{\log{\left(7 \right)}}}}$$
simplify
$$3 - x = 16$$
$$- x = 13$$
$$x = -13$$