Here's the question :
38500*1.05^x = f(x)
40000+2000x = g(x)
What is x when g(x) = f(x)
so.... 38500(1.05)^x= 40000 +2000x, can u just show the step by step solution to this. I know the answers, so don't just post the answers.
Here's the question :
38500*1.05^x = f(x)
40000+2000x = g(x)
What is x when g(x) = f(x)
so.... 38500(1.05)^x= 40000 +2000x, can u just show the step by step solution to this. I know the answers, so don't just post the answers.
You will have to use some kind of numeric root-finding or graphical technique, as algebraically solving for x is not possible here, at least not without invoking:
Lambert W function - Wikipedia, the free encyclopedia
A method I used to use when all I had was a simple calculator and hadn't learned any calculus would be to begin with:
$\displaystyle 38500(1.05)^x= 40000+2000x$
Divide through by 500:
$\displaystyle 77(1.05)^x= 80+4x$
Take a guess, observing that $\displaystyle x=0$ is fairly close but $\displaystyle x$ need to either get larger so the exponential can "catch up" or get smaller so that the linear function can catch up.
Let's try for the positive root first, and try $\displaystyle x=1$. As long as the left side is smaller we need to increase $\displaystyle x$.
$\displaystyle x=1\to80.85=84$
$\displaystyle x=2\to84.8925=88$
$\displaystyle x=3\to89.137125=92$
$\displaystyle x=4\to93.59398125=96$
$\displaystyle x=5\to98.2736803125=100$
$\displaystyle x=6\to103.187364328=104$
$\displaystyle x=7\to108.346732545=108$
At this point we know the root is between 6 and 7, and closer to 7, so let's try:
$\displaystyle x=6.75\to107.033196071=107$
$\displaystyle x=6.70\to106.772405939=106.8$
$\displaystyle x=6.725\to106.90272148=106.9$
$\displaystyle x=6.7225\to106.889682772=106.89$
$\displaystyle x=6.72275\to106.890986571=106.891$
We could carry out this process until the limit of accuracy of our calculator is reached.
A similar method can be used to find the negative root.
If you have a graphing calculator, your could use the numeric solve feature or graph the two functions and zoom in to the intersections to get a good approximation of the $\displaystyle x$-coordinate of the intersections.