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3 years ago

Write the logarithmic equation in exponential form log4 (1/4)^3 =-3

Mathematics

2 answers:

Sergeu [11.5K]3 years ago

8 0

The equation in exponentail form -->. 4^-3 = (1/4)^3 1/4^3 = 1/64 1/64 = 1/64

mr_godi [17]3 years ago

5 0

Remember
log(x^y)=ylog(x)
$log_xy=m$ means $x^m=y$
also
x^-m=1/(x^m)
so

$log_4(1/4)^3=-3$ means $4^{-3}=(1/4)^3$ or1/(4^3)=(1/4)^3=1/64=1/64

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Solve sin 0 + 1 = cos20 on the interval 0 ≤ 0 < 2pi. Show work please!

yan [13]

Answer:

$\theta=\frac{\pi}{2},\frac{3\pi}{2}\frac{2\pi}{3}\frac{4\pi}{3}$

Step-by-step explanation:

You need 2 things in order to solve this equation: a trig identity sheet and a unit circle.

You will find when you look on your trig identity sheet that

$cos(2\theta)=1-2sin^2(\theta)$

so we will make that replacement, getting everything in terms of sin:

$sin(\theta)+1=1-2sin^2(\theta)$

Now we will get everything on one side of the equals sign, set it equal to 0, and solve it:

$2sin^2(\theta)+sin(\theta)=0$

We can factor out the sin(theta), since it's common in both terms:

$sin(\theta)(2sin(\theta)+1)=0$

Because of the Zero Product Property, either

$sin(\theta)=0$ or

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Look at the unit circle and find which values of theta have a sin ratio of 0 in the interval from 0 to 2pi. They are:

$\theta=\frac{\pi}{2},\frac{3\pi}{2}$

The next equation needs to first be solved for sin(theta):

$2sin(\theta)+1=0$ so

$2sin(\theta)=-1$ and

$sin(\theta)=-\frac{1}{2}$

Go back to your unit circle and find the values of theta where the sin is -1/2 in the interval. They are:

$\theta=\frac{2\pi}{3},\frac{4\pi}{3}$

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