Mister Exam
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How to use it?
- Equation:
- Equation log1:7(7π₯β3)=log1(5π₯+11)
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Identical expressions
- log1: seven (7π₯β three)=log1(5π₯+ eleven)
- logarithm of 1:7(7π₯β3) equally logarithm of 1(5π₯ plus 11)
- logarithm of 1: seven (7π₯β three) equally logarithm of 1(5π₯ plus eleven)
- log1:77π₯β3=log15π₯+11
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Similar expressions
- log1:7(7π₯β3)=log1(5π₯-11)
log1:7(7π₯β3)=log1(5π₯+11) equation
The teacher will be very surprised to see your correct solution π
The solution
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log(1) log(5*x + 11) ------*(7*x - 3) = ------------- 7 log(1)
$$\frac{\log{\left(1 \right)}}{7} \left(7 x - 3\right) = \frac{\log{\left(5 x + 11 \right)}}{\log{\left(1 \right)}}$$
Detail solution
Given the equation
$$\frac{\log{\left(1 \right)}}{7} \left(7 x - 3\right) = \frac{\log{\left(5 x + 11 \right)}}{\log{\left(1 \right)}}$$
Transfer the right side of the equation left part with negative sign
$$\tilde{\infty} \log{\left(5 x + 11 \right)} = 0$$
Let's divide both parts of the equation by the multiplier of log =Β±oo
$$\log{\left(5 x + 11 \right)} = 0$$
This equation is of the form:
By definition log
then
$$5 x + 11 = e^{\frac{0}{\tilde{\infty}}}$$
simplify
$$5 x + 11 = 1$$
$$5 x = -10$$
$$x = -2$$
$$\frac{\log{\left(1 \right)}}{7} \left(7 x - 3\right) = \frac{\log{\left(5 x + 11 \right)}}{\log{\left(1 \right)}}$$
Transfer the right side of the equation left part with negative sign
$$\tilde{\infty} \log{\left(5 x + 11 \right)} = 0$$
Let's divide both parts of the equation by the multiplier of log =Β±oo
$$\log{\left(5 x + 11 \right)} = 0$$
This equation is of the form:
log(v)=p
By definition log
v=e^p
then
$$5 x + 11 = e^{\frac{0}{\tilde{\infty}}}$$
simplify
$$5 x + 11 = 1$$
$$5 x = -10$$
$$x = -2$$