Respuesta :

merlynthewhizz
given the exponential functions [tex]g(x)= 5^{x} [/tex] and [tex]h(x)= 5^{-x} [/tex]

[tex]k(x)=(g-h)(x)[/tex]
[tex]k(x)=g(x)-h(x) [/tex]
[tex]k(x)= 5^{x} - 5^{-x} [/tex]
[tex]k(x)= 5^{x} - \frac{1}{ 5^{x} } [/tex]
[tex]k(x)= \frac{ 5^{x} 5^{x} }{ 5^{x} } - \frac{1}{ 5^{x} } [/tex]
[tex]k(x)= \frac{ 5^{2x} }{ 5^{x} } - \frac{1}{ 5^{x} } [/tex]
[tex]k(x)= \frac{ 5^{2x}-1 }{ 5^{x} } [/tex]

Answer:

The value of [tex]k(x)=\frac{5^{2x}-1}{5^x}[/tex]

Step-by-step explanation:

We have given two function [tex]g(x)=5^x\text{and}h(x)=5^{-x}[/tex]

We have to find k(x)=(g-h)(x)

[tex]k(x)=g(x)-h(x)[/tex]           (1)

We will substitute the values in equation (1) we will get

[tex]k(x)=5^x-(5^{-x})[/tex]

Now, open the parenthesis on right hand side of equation we will get

[tex]k(x)=5^x-5^{-x}[/tex]

Using [tex]x^{-a}=\frac{1}{x^a}[/tex]

[tex]k(x)=5^x-\frac{1}{5^x}[/tex]

Now, taking LCM which is [tex]5^x[/tex] we will get after simplification

[tex]k(x)=\frac{5^{2x}-1}{5^x}[/tex]

Hence, the value of [tex]k(x)=\frac{5^{2x}-1}{5^x}[/tex]

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