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Inverse Exponential - logarithm

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Mathematics

10th - 12th Grade

CCSS covered

Used 40+ times

Inverse Exponential - logarithm
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4 questions

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1.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the inverse equation for the following function? f(x)=2(3)xf\left(x\right)=2\left(3\right)^x  

 f1(x)=log3(x2)f^{-1}\left(x\right)=\log_3\left(\frac{x}{2}\right)  

 f1(x)=log2(x3)f^{-1}\left(x\right)=\log_2\left(\frac{x}{3}\right)  

 f1(x)=log32xf^{-1}\left(x\right)=\log_32x  

 f1(x)=log23xf^{-1}\left(x\right)=\log_23x  

Tags

CCSS.HSF-BF.B.4A

2.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the inverse of the function  g(x)=2(3)(4x)g\left(x\right)=2\left(3\right)^{\left(4x\right)}  

 g1(x)=log2(x3)4g^{-1}\left(x\right)=\frac{\log_2\left(\frac{x}{3}\right)}{4}  

 g1(x)=log3(x2)4g^{-1}\left(x\right)=\frac{\log_3\left(\frac{x}{2}\right)}{4}  

 g1(x)=log3(x4)2g^{-1}\left(x\right)=\frac{\log_3\left(\frac{x}{4}\right)}{2}  

 g1(x)=log2(x4)3g^{-1}\left(x\right)=\frac{\log_2\left(\frac{x}{4}\right)}{3}  

Tags

CCSS.HSF-BF.B.4A

3.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the inverse of the function  h(x)= log3(x4)5h\left(x\right)=\ \log_3\left(\frac{x}{4}\right)-5  

 h1(x) = 4(3)(x+5)h^{-1}\left(x\right)\ =\ 4\left(3\right)^{\left(x+5\right)}  

 h1(x)=5(3)(x4)h^{-1}\left(x\right)=5\left(3\right)^{\left(\frac{x}{4}\right)}  

 h1(x)=4(3)(x5)h^{-1}\left(x\right)=4\left(3\right)^{\left(x-5\right)}  

 h1(x)=3(4)(x+5)h^{-1}\left(x\right)=3\left(4\right)^{\left(x+5\right)}  

Tags

CCSS.HSF-BF.B.4A

4.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

In a science experiment a scientist initially observes that there are 120 bacteria cells. One hour later the scientist observes that the cells have increased by 5%. The scientist models the growth of the number of cells with the function  b(h)=120(1.05)hb\left(h\right)=120\left(1.05\right)^h  , where b(x) is the number of bacteria cells and h is the number of hours. What inverse equation can predict the number of hours based on the total number of bacteria?

 b1(h)=log1.05(h120)b^{-1}\left(h\right)=\log_{1.05}\left(\frac{h}{120}\right)  

 b1(h)=log1.05(120(h))b^{-1}\left(h\right)=\log_{1.05}\left(\frac{120}{\left(h\right)}\right)  

 b1(h)=log120(h1.05)b^{-1}\left(h\right)=\log_{120}\left(\frac{h}{1.05}\right)  

 b1(h)=log120(1.05h)b^{-1}\left(h\right)=\log_{120}\left(\frac{1.05}{h}\right)  

Tags

CCSS.HSF.BF.B.5

CCSS.HSF.LE.A.4

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