# (Aust)Yr11 logs assignment (Due nxt Thur)

• October 19th 2007, 04:35 PM
mathsdunce000
(Aust)Yr11 logs assignment (Due nxt Thur)
Hi everyone, thank you for your help. First of all, i dont expect a straightforward answer, more like hints and clues to work it out myself so i can actually learn something.

Eq1) 5x2^(x+1) - 12 = 0
eq2) 3x2^x = 4x3^(-x)+7

a) Which one of the above can NOT be solved using logs?

b) Solve the equation that is the answer to (a) using a G.C. Give a sketch of your solution.

c) Solve the other equation using logs and give answer in log form.

[by the way, how do you type in the equations as you would actually write them?]

Construct an exponential equation that follows this graph :

http://img484.imageshack.us/img484/2...expohl0.th.png

Written in the form y = a X b^(cx) + d
= a X (b^c)^x + d
a X k^x + d

Thank you!
• October 19th 2007, 09:15 PM
Soroban
Hello, mathsdunce000!

Quote:

$\begin{array}{cc}\text{Eq. 1} & 5\!\cdot\!2^{x+1} - 12 \:= \:0\\
\text{Eq. 2} & 3\!\cdot\!2^x \:= \:4\!\cdot\!3^{-x}+7 \end{array}$

a) Which one of the above can NOT be solved using logs?

Equation #2

Quote:

b) Solve the equation that is the answer to (a) using a G.C.
Give a sketch of your solution.

We have: . $5\!\cdot\!2^{x+1} - 12 \:=\:0\quad\Rightarrow\quad 2^{x+1} \:=\:\frac{12}{5}\:=\:2.4$

Take logs of both sides: . $\ln\left(2^{x+1}\right) \:=\:\ln(2.4)\quad\Rightarrow\quad (x+1)\ln2\:=\:\ln(2.4)$

Then: . $x+1\:=\:\frac{\ln(2.4)}{\ln(2)}\quad\Rightarrow\qu ad x \;=\;\frac{\ln(2.4)}{\ln(2)} -1 \;\approx\;0.263034406$

Quote:

c) Solve the other equation using logs and give answer in log form. . ??
Didn't we say: ."Equation 2 cannot be solved using logs" ?

• October 19th 2007, 10:21 PM
mathsdunce000
Hi, thanks very much for your time and effort! b) was solving equation 2 (the answer to question a)) then solving that using a GC. That was the question i was stumped on.

so i was just wondering how would i use a GC to solve equation 2.
c) was what you did for b).

Thanks alot!
• October 20th 2007, 05:18 AM
topsquark
Quote:

Originally Posted by mathsdunce000
eq2) 3x2^x = 4x3^(-x)+7

Quote:

Originally Posted by mathsdunce000
Hi, thanks very much for your time and effort! b) was solving equation 2 (the answer to question a)) then solving that using a GC. That was the question i was stumped on.

so i was just wondering how would i use a GC to solve equation 2.
c) was what you did for b).

Thanks alot!

You start by multiplying both sides by $3^x$. This gives you:
$(3 \cdot 2^x)3^x = (4 \cdot 3^{-x}+7)3^x$

$2^x3^{x + 1} = 4 \cdot 1 + 7 \cdot 3^x$

$2^x3^{x + 1} = 4 + 7 \cdot 3^x$

Now you can solve this using logs.

-Dan

PS By the way. When writing an expression such as $3 \cdot 2^x$ it is confusing to write "3x2^x" as the x's imply $(3x)2^x$, which is not what you wanted. In this case I would write this as "(3)(2^x)" or something so there is no confusion.
• October 20th 2007, 04:45 PM
mathsdunce000
thank you soroban and squark for your help! Any ideas on the graph?
• October 20th 2007, 05:15 PM
topsquark
Quote:

Originally Posted by mathsdunce000
Construct an exponential equation that follows this graph :

http://img484.imageshack.us/img484/2...expohl0.th.png

Written in the form y = a X b^(cx) + d
= a X (b^c)^x + d
a X k^x + d

$y = a \cdot b^{cx} + d$

You know that as x goes to infinity that y goes to 12. This tells us two things:
First that the sign of c is negative, else the function blows up as x goes to infinity.
Second that d = 12. (The limit of x going to infinity of the exponential with a negative sign goes to 0.)

So let's rewrite this as
$y = a \cdot b^{-cx} + 12$
(I'm redefining c here to be positive.)

You have 3 points on your curve:
(0, 62), (5, 26.519), and (12, 14.57)

Plug each of these into your equation. The first gives you a value for a:
$62 = a \cdot b^{-c \cdot 0} + 12$

$62 = a + 12 \implies a = 50$

So we have
$y = 50 \cdot b^{-cx} + 12$

The second and third points give:
$26.519 = 50 \cdot b^{-5c} + 12 \implies b^{-5c} = \frac{26.519}{50} = 0.29038$
and
$14.57 = 50 \cdot b^{-12c} \implies b^{-12c} = \frac{14.57}{50} = 0.0514$

How do you solve these?
Well, we have a problem:
$(b^{-5c})^{-1/5} = b^c = \frac{1}{(0.29038)^{1/5}} = 1.81054$
and
$(b^{-12c})^{-1/12} = b^c = \frac{1}{(0.0514)^{1/12}} = 1.28062$

This is an inconsistency. Perhaps this is due to estimating the points that are graphed? Either way, this system cannot be solved as it is.

-Dan