Which is the graph of f(x) = log3x?.
1 answer:
Logarithm functions are the inverses of the exponential function the graph of the given function is attached in the given image.
The points for the given function are (1,0) (3,1) and (9,2)
Given that
Function;
We have to determine
The graph of the function f(x).
According to the question
The logarithmic function is the inverse of the exponential function.
Function;
To plot the graph of the given function we have to find the asymptotes;
Vertical asymptote at x =0
The point at x= 1 is,
The point at x= 3 is,
The point at x= 9 is,
Hence, the points for the given function are (1,0) (3,1) and (9,2).
To know more about the Exponential function given below.
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Answer:
99% confidence interval: (0.97816,1.03184)
Step-by-step explanation:
We are given the following data set:
1.01, 0.97, 1.03, 1.04, 0.99, 0.98, 0.99, 1.01, 1.03
Formula:
where are data points,
is the mean and n is the number of observations.
Sum of squares of differences = 0.00001975308642 + 0.001264197531 + 0.000597530864 + 0.001186419753 + 0.0002419753088, + 0.00065308642 + 0.0002419753088 + 0.00001975308642 + 0.000597530864 = 0.0048
Confidence interval:
Putting the values, we get,
First, we should know exponential formula, for example exp (LnA) =A, for all A>0, and as for logarythm, LnA^p=pLnA, for all A>0, and for all value of p
so y=3^x=exp[Ln(3^x)]=exp(xLn3), let's search the graph of f(x)=exp(xLn3) f is defined in R, and its teminals limits are 0, and + infinity, f' (x) = Ln3.exp(xLn3), it is positif for all value of x, the function is strictly increasing, f(0)=Ln3, please look at the figure 1
the same method with y=2(4.5)^x, we find y=exp[Ln (2(4.5)^x)]=
exp[Ln (2) +Ln(4.5)^x)], because (LnAB=LnA + LnB), so y=2.exp[Ln(4.5)^x)]=2.exp(x[Ln(4.5)]), which domain is R, nd its teminals limits are 0, and + infinity, f' (x)=2.Ln(4.5) exp(x[Ln(4.5)]), which is positif for all value of x, the function is strictly increasing,and then f(0)=2Ln(4.5)
please look the figure 2
so y=3^x=exp[Ln(3^x)]=exp(xLn3), let's search the graph of f(x)=exp(xLn3) f is defined in R, and its teminals limits are 0, and + infinity, f' (x) = Ln3.exp(xLn3), it is positif for all value of x, the function is strictly increasing, f(0)=Ln3, please look at the figure 1
the same method with y=2(4.5)^x, we find y=exp[Ln (2(4.5)^x)]=
exp[Ln (2) +Ln(4.5)^x)], because (LnAB=LnA + LnB), so y=2.exp[Ln(4.5)^x)]=2.exp(x[Ln(4.5)]), which domain is R, nd its teminals limits are 0, and + infinity, f' (x)=2.Ln(4.5) exp(x[Ln(4.5)]), which is positif for all value of x, the function is strictly increasing,and then f(0)=2Ln(4.5)
please look the figure 2