60*80-10*16 if it's regular area and not surface area
Answer:
![\large\boxed{1.\ f^{-1}(x)=4\log(x\sqrt[4]2)}\\\\\boxed{2.\ f^{-1}(x)=\log(x^5+5)}\\\\\boxed{3.\ f^{-1}(x)=\sqrt{4^{x-1}}}](https://tex.z-dn.net/?f=%5Clarge%5Cboxed%7B1.%5C%20f%5E%7B-1%7D%28x%29%3D4%5Clog%28x%5Csqrt%5B4%5D2%29%7D%5C%5C%5C%5C%5Cboxed%7B2.%5C%20f%5E%7B-1%7D%28x%29%3D%5Clog%28x%5E5%2B5%29%7D%5C%5C%5C%5C%5Cboxed%7B3.%5C%20f%5E%7B-1%7D%28x%29%3D%5Csqrt%7B4%5E%7Bx-1%7D%7D%7D)
Step-by-step explanation:
![\log_55^{\frac{1}{4}y}=\log_5\left(2^\frac{1}{4}x\right)\qquad\text{use}\ a^\frac{1}{n}=\sqrt[n]{a}\\\\\dfrac{1}{4}y=\log(x\sqrt[4]2)\qquad\text{multiply both sides by 4}\\\\y=4\log(x\sqrt[4]2)](https://tex.z-dn.net/?f=%5Clog_55%5E%7B%5Cfrac%7B1%7D%7B4%7Dy%7D%3D%5Clog_5%5Cleft%282%5E%5Cfrac%7B1%7D%7B4%7Dx%5Cright%29%5Cqquad%5Ctext%7Buse%7D%5C%20a%5E%5Cfrac%7B1%7D%7Bn%7D%3D%5Csqrt%5Bn%5D%7Ba%7D%5C%5C%5C%5C%5Cdfrac%7B1%7D%7B4%7Dy%3D%5Clog%28x%5Csqrt%5B4%5D2%29%5Cqquad%5Ctext%7Bmultiply%20both%20sides%20by%204%7D%5C%5C%5C%5Cy%3D4%5Clog%28x%5Csqrt%5B4%5D2%29)
![--------------------------\\2.\\y=(10^x-5)^\frac{1}{5}\\\\\text{Exchange x and y. Solve for y:}\\\\(10^y-5)^\frac{1}{5}=x\qquad\text{5 power of both sides}\\\\\bigg[(10^y-5)^\frac{1}{5}\bigg]^5=x^5\qquad\text{use}\ (a^n)^m=a^{nm}\\\\(10^y-5)^{\frac{1}{5}\cdot5}=x^5\\\\10^y-5=x^5\qquad\text{add 5 to both sides}\\\\10^y=x^5+5\qquad\log\ \text{of both sides}\\\\\log10^y=\log(x^5+5)\Rightarrow y=\log(x^5+5)](https://tex.z-dn.net/?f=--------------------------%5C%5C2.%5C%5Cy%3D%2810%5Ex-5%29%5E%5Cfrac%7B1%7D%7B5%7D%5C%5C%5C%5C%5Ctext%7BExchange%20x%20and%20y.%20Solve%20for%20y%3A%7D%5C%5C%5C%5C%2810%5Ey-5%29%5E%5Cfrac%7B1%7D%7B5%7D%3Dx%5Cqquad%5Ctext%7B5%20power%20of%20both%20sides%7D%5C%5C%5C%5C%5Cbigg%5B%2810%5Ey-5%29%5E%5Cfrac%7B1%7D%7B5%7D%5Cbigg%5D%5E5%3Dx%5E5%5Cqquad%5Ctext%7Buse%7D%5C%20%28a%5En%29%5Em%3Da%5E%7Bnm%7D%5C%5C%5C%5C%2810%5Ey-5%29%5E%7B%5Cfrac%7B1%7D%7B5%7D%5Ccdot5%7D%3Dx%5E5%5C%5C%5C%5C10%5Ey-5%3Dx%5E5%5Cqquad%5Ctext%7Badd%205%20to%20both%20sides%7D%5C%5C%5C%5C10%5Ey%3Dx%5E5%2B5%5Cqquad%5Clog%5C%20%5Ctext%7Bof%20both%20sides%7D%5C%5C%5C%5C%5Clog10%5Ey%3D%5Clog%28x%5E5%2B5%29%5CRightarrow%20y%3D%5Clog%28x%5E5%2B5%29)
Answer:
64
Step-by-step explanation:
2*2*2*2*2*2
2 multiply 2 multiply 2 multiply 2 multiply 2 multiply 2.
So you follow the steps and i hope this will be easy for you .
21−=2(2−)=2cos(−1)+2 sin(−1)
−1+2=−1(2)=−1(cos2+sin2)=cos2+ sin2
Is the above the correct way to write 21− and −1+2 in the form +? I wasn't sure if I could change Euler's formula to =cos()+sin(), where is a constant.
complex-numbers
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edited Mar 6 '17 at 4:38
Richard Ambler
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asked Mar 6 '17 at 3:34
14wml
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1 Answer
1
No. It is not true that =cos()+sin(). Notice that
1=1≠cos()+sin(),
for example consider this at =0.
As a hint for figuring this out, notice that
+=ln(+)
then recall your rules for logarithms to get this to the form (+)ln().
