Mister Exam
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How to use it?
- Equation:
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Equation 2tgx–9ctgx+3=0
- Equation 2log16(2x−5)=2
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Equation 16+6x=5(1-2x)-13
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Identical expressions
- two log16(2x− five)=2
- 2 logarithm of 16(2x−5) equally 2
- two logarithm of 16(2x− five) equally 2
- 2log162x−5=2
2log16(2x−5)=2 equation
The teacher will be very surprised to see your correct solution 😉
The solution
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[src]
log(2*x - 5)
2*------------ = 2
log(16)
$$2 \frac{\log{\left(2 x - 5 \right)}}{\log{\left(16 \right)}} = 2$$
Detail solution
Given the equation
$$2 \frac{\log{\left(2 x - 5 \right)}}{\log{\left(16 \right)}} = 2$$
$$\frac{2 \log{\left(2 x - 5 \right)}}{\log{\left(16 \right)}} = 2$$
Let's divide both parts of the equation by the multiplier of log =2/log(16)
$$\log{\left(2 x - 5 \right)} = \log{\left(16 \right)}$$
This equation is of the form:
By definition log
then
$$2 x - 5 = e^{\frac{2}{2 \frac{1}{\log{\left(16 \right)}}}}$$
simplify
$$2 x - 5 = 16$$
$$2 x = 21$$
$$x = \frac{21}{2}$$
$$2 \frac{\log{\left(2 x - 5 \right)}}{\log{\left(16 \right)}} = 2$$
$$\frac{2 \log{\left(2 x - 5 \right)}}{\log{\left(16 \right)}} = 2$$
Let's divide both parts of the equation by the multiplier of log =2/log(16)
$$\log{\left(2 x - 5 \right)} = \log{\left(16 \right)}$$
This equation is of the form:
log(v)=p
By definition log
v=e^p
then
$$2 x - 5 = e^{\frac{2}{2 \frac{1}{\log{\left(16 \right)}}}}$$
simplify
$$2 x - 5 = 16$$
$$2 x = 21$$
$$x = \frac{21}{2}$$
Sum and product of roots
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sum
21/2
$$\frac{21}{2}$$
=
21/2
$$\frac{21}{2}$$
product
21/2
$$\frac{21}{2}$$
=
21/2
$$\frac{21}{2}$$
21/2