Question

f(x)=14000(1-0.09)^x when will it be 7000

Answer 1

To find out when \( f(x) = 7000 \), we start with the given equation:

\[
f(x) = 14000(1 - 0.09)^x
\]

Setting \( f(x) = 7000 \):

\[
7000 = 14000(1 - 0.09)^x
\]

Dividing both sides by 14000:

\[
\frac{7000}{14000} = (1 - 0.09)^x
\]

This simplifies to:

\[
0.5 = (0.91)^x
\]

Next, we take the logarithm of both sides. We can use the natural logarithm (ln), but any logarithm will work:

\[
\ln(0.5) = \ln((0.91)^x)
\]

Using the power property of logarithms, this becomes:

\[
\ln(0.5) = x \ln(0.91)
\]

Now, solve for \( x \):

\[
x = \frac{\ln(0.5)}{\ln(0.91)}
\]

Calculating the values:

\[
\ln(0.5) \approx -0.693147
\]
\[
\ln(0.91) \approx -0.094310
\]

Now substitute these values:

\[
x \approx \frac{-0.693147}{-0.094310} \approx 7.36
\]

Since \( x \) typically represents time in discrete terms (like years), we can round up, thus it will take approximately 8 units of time (depending on the context of \( x \), e.g., years, months, etc.) for \( f(x) \) to reach 7000.