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3 years ago

10^(-10log3) How to solve this???

1 answer:

4 0

$a^{log_ab}=b\\and\\log_ab^c=c\cdot log_ab\\\\\\10^{-10log3}=10^{log3^{-10}}=3^{-10}\\\\\\3^{-10}=\frac{1}{3^{10}}=\frac{1}{59\ 049}$

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3 years ago

Calculating cos-1 ( help is gladly appreciated :) )

Alekssandra [29.7K]

Answer:

$\frac{3\pi}{4}$

(Assuming you want your answer in radians)

If you want the answer in degrees just multiply your answer in radians by $\frac{180^\circ}{\pi}$ giving you:

$\frac{3\pi}{4} \cdot \frac{180^\circ}{\pi}=\frac{3(180)}{4}=135^{\circ}$ .

We can do this since $\pi \text{ rad }=180^\circ$ (half the circumference of the unit circle is equivalent to 180 degree rotation).

Step-by-step explanation:

$\cos^{-1}(x)$ is going to output an angle measurement in $[0,\pi]$ .

So we are looking to solve the following equation in that interval:

$\cos(x)=-\frac{\sqrt{2}}{2}$ .

This happens in the second quadrant on the given interval.

The solution to the equation is $\frac{3\pi}{4}$ .

So we are saying that $\cos(\frac{3\pi}{4})=\frac{-\sqrt{2}}{2}$ implies $\cos^{-1}(\frac{-\sqrt{2}}{2})=\frac{3\pi}{4}$ since $\frac{3\pi}{4} \in [0,\pi]$ .

Answer is $\frac{3\pi}{4}$ .

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3 years ago