Question

if f(x) = r to the power x where r > 0 and if f(6) = 125, what is the value of f(7) ?

help fpr gods sake

Answer 1

Answer:

$f(7)=125\sqrt{5}$

Step-by-step explanation:

Given function:

$f(x)=r^x, \quad r > 0$

If f(6) = 125, substitute x = 6 into the function and equate it to 125:

$\implies r^6=125$

Rewrite 125 as 5³:

$\implies r^6=5^3$

Cube root both sides of the equation:

$\implies \sqrt[3]{r^6}=\sqrt[3]{5^3}$

$\textsf{Apply exponent rule} \quad \sqrt[n]{a^b}=a^{\frac{b}{n}}:$

$\implies r^{\frac{6}{3}}=5^{\frac{3}{3}}$

$\implies r^{2}=5^1$

$\implies r^{2}=5$

Square root both sides of the equation:

$\implies \sqrt{r^2}=\sqrt{5}$

$\implies r=\pm \sqrt{5}$

As r > 0 then r = √5

Therefore the function is:

$f(x)=\left(\sqrt{5} \right)^x$

To calculate f(7), substitute x = 7 into the found function:

$\implies f(7)=\left(\sqrt{5} \right)^7$

$\textsf{Apply exponent rule} \quad \sqrt[n]{a}=a^{\frac{1}{n}}:$

$\implies f(7)=(5^{\frac{1}{2}})^7$

$\textsf{Apply exponent rule} \quad (a^b)^c=a^{bc}:$

$\implies f(7)=5^{\frac{7}{2}}$

Rewrite ⁷/₂ as 3 + ¹/₂ :

$\implies f(7)=5^{\left(3+\frac{1}{2}\right)}$

$\textsf{Apply exponent rule} \quad a^{b+c}=a^b \cdot a^c:$

$\implies f(7)=5^3 \cdot 5^{\frac{1}{2}}$

Simplify:

$\implies f(7)=125 \sqrt{5}$