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enter image description here

  1. For this, if i take the log, I know that $\ln (4) > \ln(3)$ The question doesn't specify if $x>0$ but I will assume so. So I have

$-x\ln(4) < -x\ln(3)$ So $g(x)=3^{-x}$ is the "bigger graph." my question is, how do i know which is the "bigger" graph but just looking at the picture?

  1. enter image description here

I know I have to set the equations equal to each other, so I have:

$3^{2x}-3^x =20$

$(3^x)^2 - 3^x = 20$

$3^x(3^x -1) = 20$

Confused on where to go from here...

user130306
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2 Answers2

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Note that $$f(x)=\left(\frac{1}{4}\right)^x\space\text{and}\space g(x)=\left(\frac{1}{3}\right)^x$$

Since successive powers of $\frac{1}{4}$ are smaller than successive powers of $\frac{1}{3},$ $g(x)$ should be the bigger or taller graph for $x>0$, and this is the blue graph. By taller we mean the graph that lies above the other in terms of height.

For the second question, $$3^{2x}-3^x=20\iff (3^x)^2-(3^x)-20=0$$

Now let $u=3^x$ so that $$(3^x)^2-(3^x)-20=0\iff u^2-u-20=0$$

Can you take it from here?

coreyman317
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  • thanks so much! i can accept your answer in 6 minutes, but just another question on part 1: are you saying that for every x, the y-value should be greater for it to be the "bigger" graph? because for the red graph, on the right side it is true but not on the LHS of the graph – user130306 May 08 '19 at 02:52
  • im just still a little confused about what "bigger" means – user130306 May 08 '19 at 02:52
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For the first graph, the red one is $y=4^{-x}$ because for positive values of $x$ it is less than $3^{-x}$ and for negative values of $x$ it is larger than $3^{-x}$

For the second graph you have $t^2-t-20=0$ where $t=3^x$

The positive value of $t$ is $5$, that is $3^x=5$ or $x=\log _3 5$

Mohammad Riazi-Kermani
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