I’ll give the Brainliest to who answers my question and gives an reasonable explanation.
1 answer:
You might be interested in
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