Answer to Question #350228 in Algebra for owamiportia.22@gma

Question

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

Answer 1

4.1

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

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

$$g(x)=\log_4(x+2)+\log_4(3)-\log_4(5)-\log_4(2x-3)$$ $$x+2>0=>x>-2$$ $$2x-3>0=>x>\dfrac{3}{2}$$

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

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

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

4.2

$$f(x)=12$$

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

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

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

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

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

$$x_2=\dfrac{\ln(\dfrac{9+\sqrt{77}}{2})}{\ln3}$$

4.3

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

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

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

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

$$x=3$$

4.4

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

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

$$(e^x-3)(e^x-5)=0$$

$$e^x-3=0\ or\ e^x-5=0$$

$$x_1=\ln 3, x_2=\ln 5$$