Question #350228

Suppose the functions f,g and h are defined by f(x)=3^x+3^-x+3 and g(x)=log5(x+2)+log5(3)-log4(2x-3) and h(x)=e^x+15e^-x respectively 4.1 write down the sets Df,DG and Dh that represent the domains of f g and h respectively. 4.2 solve the equation f(x)=12 for x. 4.3 solve the inequality g(x)= 0 for x. 4.4solve h(x)=8 for x. Leave the answer in terms of ln, where necessary


1
Expert's answer
2022-06-13T15:32:24-0400

4.1


f(x)=3x+3x+3f(x)=3^x+3^{-x}+3

Df:(,)Df: (-\infin, \infin)



g(x)=log4(x+2)+log4(3)log4(5)log4(2x3)g(x)=\log_4(x+2)+\log_4(3)-\log_4(5)-\log_4(2x-3)x+2>0=>x>2x+2>0=>x>-22x3>0=>x>322x-3>0=>x>\dfrac{3}{2}

Dg:(32,)Dg:(\dfrac{3}{2}, \infin)




h(x)=ex+15exh(x)=e^x+15e^{-x}

Dh:(,)Dh: (-\infin, \infin)


4.2


f(x)=12f(x)=12

3x+3x+3=123^x+3^{-x}+3=12

(3x)29(3x)+1=0(3^x)^2-9(3^x)+1=0

Determinant=(9)24(1)(1)=77Determinant=(-9)^2-4(1)(1)=77

3x=9±772(1)3^x=\dfrac{9\pm\sqrt{77}}{2(1)}

x1=ln(9772)ln3,x_1=\dfrac{\ln(\dfrac{9-\sqrt{77}}{2})}{\ln3},

x2=ln(9+772)ln3x_2=\dfrac{\ln(\dfrac{9+\sqrt{77}}{2})}{\ln3}

4.3


log4(x+2)+log4(3)log4(5)log4(2x3)=0\log_4(x+2)+\log_4(3)-\log_4(5)-\log_4(2x-3)=0

log43(x+2)5(2x3)=0\log_4\dfrac{3(x+2)}{5(2x-3)}=0

3(x+2)5(2x3)=1\dfrac{3(x+2)}{5(2x-3)}=1

3x+6=10x153x+6=10x-15

x=3x=3

4.4


ex+15ex=8e^x+15e^{-x}=8

(ex)28(ex)+15=0(e^x)^2-8(e^x)+15=0

(ex3)(ex5)=0(e^x-3)(e^x-5)=0

ex3=0 or ex5=0e^x-3=0\ or\ e^x-5=0

x1=ln3,x2=ln5x_1=\ln 3, x_2=\ln 5


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