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3 years ago

Can a TI-84 Plus simplify radicals?

Mathematics

1 answer:

Mnenie [13.5K]3 years ago

4 0

TI-84 Plus does not simplify radicals

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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

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

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

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

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

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

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

$x=4$

Therefore, the solution is

$x=4$

6 0

3 years ago

Read 2 more answers

Find the differnce of the following sequence: -7, -9 ,-11

Slav-nsk [51]

Add -2 each time
-7+-2=-9
-9+-2=-11

8 0

3 years ago

1. Solve the system by using a table. 3y=4x+7 , -4x-4y=28. These are the possible solutions: A) (–4, –3), B) (–3, –4), C) (4, 3)

melisa1 [442]

The first question:
3y=4x+7 (1)
-4x-4y=28 (2)

Let's plug in x = -4, -3, 3, and 4 and see the y-values :)

I have attached the table since it's hard to make a table with text :P
As you can see, when x = -4, that is when the y-values are equal. That means that is the solution to the system of equations. Your answer is A) (-4, -3).

The second question:
-3x-y=-10 (1)
4x-4y=8 (2)

When you graph both equations, you will see that they intersect at D) (3, 1).

The third question:
We need to find the lines for revenues and expenses.
To find the line for revenues, make months the x-value and revenues the y-values. And find the equation. You should get y = 4,100x + 300.
To find the line for expenses, make months the x-value and expenses the y-values. And find the equation. You should get y = 1,900x + 19,990.

Now graph both solutions and see where they intersect. They intersect at approximately (8.95, 36995)

That would be during the month of C) August.