If Ken wanted to create a function that modeled an exponential function with base of 11 and what exponents were needed to reach

specific values, how would he set up his function?
f(x) = 11x
f(x) = x11
f(x) = log11x
f(x) = logx11

Mathematics

1 answer:

Rudiy274 years ago

4 0

$\bf \textit{exponential form of a logarithm}
\\\\
log_a b=y \implies a^y= b\qquad\qquad 
% exponential notation 2nd form
a^y= b\implies log_a b=y \\\\
-------------------------------\\\\
x=11^y\implies log_{11}(x)=y\implies log_{11}(x)=f(x)$

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7.209 x 10 -4 in standard form

astraxan [27]

$\huge\textbf{Hey there!}$

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2 years ago

Bet u can’t solve this

S_A_V [24]

Answer:

Step-by-step explanation:

Assuming you're solving for p:

$m=2^{-1} *2^p(2^p-1)$

Let $y=2^p$

Now we can re-write the equation with $y$ instead of $2^p$ .

$m=\frac{1}{2} y(y-1)$

$2m=y^2-y$

$y^2-y-2m=0$

Use the quadratic formula to get:

$y = \frac{1+\sqrt{1+8m} }{2}$

$y = \frac{1-\sqrt{1+8m} }{2}$

Therefore, using natural log and log rules:

$2^p = \frac{1+\sqrt{1+8m} }{2}$ , $ln(2^p)= ln(\frac{1+\sqrt{1+8m} }{2})$ , $pln(2) = ln(\frac{1+\sqrt{1+8m} }{2})$ , $p = \frac{ln(\frac{1+\sqrt{1+8m} }{2})}{ln(2)}$

$2^p = \frac{1-\sqrt{1+8m} }{2}$ , $ln(2^p)= ln(\frac{1-\sqrt{1+8m} }{2})$ , $pln(2) = ln(\frac{1-\sqrt{1+8m} }{2})$ , $p = \frac{ln(\frac{1-\sqrt{1+8m} }{2})}{ln(2)}$