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3 years ago

Hey can you please help me posted picture of question

Mathematics

1 answer:

GalinKa [24]3 years ago

5 0

The given equation is:
ax2 + bx + c = 0
We have the resolvent is:
x = (- b +/- root (b2 - 4ac)) / (2a)
The discriminant is:
b2 - 4ac = 0
The solution will be:
x = (- b) / (2a)
Thus, the equation has a real solution.
Answer:
option B

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What is the true solution to the equation below?

Marina CMI [18]

Answer:

The solution is:

$x=4$

Step-by-step explanation:

Considering the expression

$lne^{lnx}+lne^{lnx}^{^2}=2ln8$

$\ln \left(e^{\ln \left(x\right)}\right)+\ln \left(e^{\ln \left(x\right)\cdot \:2}\right)=2\ln \left(8\right)$

$\mathrm{Apply\:log\:rule}:\quad \:log_a\left(a^b\right)=b$

$\ln \left(e^{\ln \left(x\right)}\right)=\ln \left(x\right),\:\space\ln \left(e^{\ln \left(x\right)2}\right)=\ln \left(x\right)2$

$\ln \left(x\right)+\ln \left(x\right)\cdot \:2=2\ln \left(8\right)$

$\mathrm{Add\:similar\:elements:}\:\ln \left(x\right)+2\ln \left(x\right)=3\ln \left(x\right)$

$3\ln \left(x\right)=2\ln \left(8\right)$

$\mathrm{Divide\:both\:sides\:by\:}3$

$\frac{3\ln \left(x\right)}{3}=\frac{2\ln \left(8\right)}{3}$

$\ln \left(x\right)=\frac{2\ln \left(8\right)}{3}.....A$

Solving the right side of the equation A.

$\frac{2\ln \left(8\right)}{3}$

$\ln \left(8\right):\quad 3\ln \left(2\right)$

Because

$\ln \left(8\right)$

$\mathrm{Rewrite\:}8\mathrm{\:in\:power-base\:form:}\quad 8=2^3$

⇒ $\ln \left(2^3\right)$

$\mathrm{Apply\:log\:rule}:\quad \log _a\left(x^b\right)=b\cdot \log _a\left(x\right)$

$\ln \left(2^3\right)=3\ln \left(2\right)$

$\frac{2\ln \left(8\right)}{3}=\frac{2\cdot \:3\ln \left(2\right)}{3}$

$\mathrm{Multiply\:the\:numbers:}\:2\cdot \:3=6$

$=\frac{6\ln \left(2\right)}{3}$

$\mathrm{Divide\:the\:numbers:}\:\frac{6}{3}=2$

$=2\ln \left(2\right)$

So, equation A becomes

$\ln \left(x\right)=2\ln \left(2\right)$

$\mathrm{Apply\:log\:rule}:\quad \:a\log _c\left(b\right)=\log _c\left(b^a\right)$

$=\ln \left(2^2\right)$

$=\ln \left(4\right)$

$\ln \left(x\right)=\ln \left(4\right)$

$\mathrm{Apply\:log\:rule:\:\:If}\:\log _b\left(f\left(x\right)\right)=\log _b\left(g\left(x\right)\right)\:\mathrm{then}\:f\left(x\right)=g\left(x\right)$