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mr_godi [17]
3 years ago
12

Keisha is solving the equation 7 to the power of x equals 9. Which equation shows the first step she should take?

Mathematics
1 answer:
Lerok [7]3 years ago
8 0

Answer:

The correct answers are:

Question 4: the first step Keisha should take is: log to the power of x equals log 9.

Question 5: step 4 could be n ln 2 equals ln 86

Question 6: the exact solution is x=ln(7)

Step-by-step explanation:

Ok,

Question 4: the equation that shows the first step Keisha should take is:

7^{x}=9

log(7^{x} )=log(9)

x(log(7))=log(9) (applying the properties of logarithms)

x=\frac{log(9)}{log(7)} =1.129

Solution: the first step Keisha should take is: log to the power of x equals log 9.

Question 5: Her first three steps are:

2^{n}-3=83

2^{n}-3+3=83+3 (adding 3 to both sides)

2^{n}=86

2^{n}-3+3=83+3

nln(2)=ln(86)

Solution: step 4 is: nln(2)=ln(86) (applying the properties of logarithms)

Question 6: The exact solution of 2 e to the power of x equals 14 is:

2e^{x}=14

\frac{2e^{x} }{2} =\frac{14}{2} (dividing both sides by 2)

e^{x}=7

xln(e)=ln7 (ln(e)=1)

x=ln(7)

Solution: x=ln(7)

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2 years ago
I need help with my math homework. The questions is: Find all solutions of the equation in the interval [0,2π).
Aleksandr-060686 [28]

Answer:

\frac{7\pi}{24} and \frac{31\pi}{24}

Step-by-step explanation:

\sqrt{3} \tan(x-\frac{\pi}{8})-1=0

Let's first isolate the trig function.

Add 1 one on both sides:

\sqrt{3} \tan(x-\frac{\pi}{8})=1

Divide both sides by \sqrt{3}:

\tan(x-\frac{\pi}{8})=\frac{1}{\sqrt{3}}

Now recall \tan(u)=\frac{\sin(u)}{\cos(u)}.

\frac{1}{\sqrt{3}}=\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}

or

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The first ratio I have can be found using \frac{\pi}{6} in the first rotation of the unit circle.

The second ratio I have can be found using \frac{7\pi}{6} you can see this is on the same line as the \frac{\pi}{6} so you could write \frac{7\pi}{6} as \frac{\pi}{6}+\pi.

So this means the following:

\tan(x-\frac{\pi}{8})=\frac{1}{\sqrt{3}}

is true when x-\frac{\pi}{8}=\frac{\pi}{6}+n \pi

where n is integer.

Integers are the set containing {..,-3,-2,-1,0,1,2,3,...}.

So now we have a linear equation to solve:

x-\frac{\pi}{8}=\frac{\pi}{6}+n \pi

Add \frac{\pi}{8} on both sides:

x=\frac{\pi}{6}+\frac{\pi}{8}+n \pi

Find common denominator between the first two terms on the right.

That is 24.

x=\frac{4\pi}{24}+\frac{3\pi}{24}+n \pi

x=\frac{7\pi}{24}+n \pi (So this is for all the solutions.)

Now I just notice that it said find all the solutions in the interval [0,2\pi).

So if \sqrt{3} \tan(x-\frac{\pi}{8})-1=0 and we let u=x-\frac{\pi}{8}, then solving for x gives us:

u+\frac{\pi}{8}=x ( I just added \frac{\pi}{8} on both sides.)

So recall 0\le x.

Then 0 \le u+\frac{\pi}{8}.

Subtract \frac{\pi}{8} on both sides:

-\frac{\pi}{8}\le u

Simplify:

-\frac{\pi}{8}\le u

-\frac{\pi}{8}\le u

So we want to find solutions to:

\tan(u)=\frac{1}{\sqrt{3}} with the condition:

-\frac{\pi}{8}\le u

That's just at \frac{\pi}{6} and \frac{7\pi}{6}

So now adding \frac{\pi}{8} to both gives us the solutions to:

\tan(x-\frac{\pi}{8})=\frac{1}{\sqrt{3}} in the interval:

0\le x.

The solutions we are looking for are:

\frac{\pi}{6}+\frac{\pi}{8} and \frac{7\pi}{6}+\frac{\pi}{8}

Let's simplifying:

(\frac{1}{6}+\frac{1}{8})\pi and (\frac{7}{6}+\frac{1}{8})\pi

\frac{7}{24}\pi and \frac{31}{24}\pi

\frac{7\pi}{24} and \frac{31\pi}{24}

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3 years ago
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