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

1 answer

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.