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

1. ## (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 :

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

Thank you!

2. Hello, mathsdunce000!

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

a) Which one of the above can NOT be solved using logs?
Equation #2

b) Solve the equation that is the answer to (a) using a G.C.
Give a sketch of your solution.
We have: .$\displaystyle 5\!\cdot\!2^{x+1} - 12 \:=\:0\quad\Rightarrow\quad 2^{x+1} \:=\:\frac{12}{5}\:=\:2.4$

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

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

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

3. 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!

4. Originally Posted by mathsdunce000
eq2) 3x2^x = 4x3^(-x)+7
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 $\displaystyle 3^x$. This gives you:
$\displaystyle (3 \cdot 2^x)3^x = (4 \cdot 3^{-x}+7)3^x$

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

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

Now you can solve this using logs.

-Dan

PS By the way. When writing an expression such as $\displaystyle 3 \cdot 2^x$ it is confusing to write "3x2^x" as the x's imply $\displaystyle (3x)2^x$, which is not what you wanted. In this case I would write this as "(3)(2^x)" or something so there is no confusion.

5. thank you soroban and squark for your help! Any ideas on the graph?

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

Written in the form y = a X b^(cx) + d
= a X (b^c)^x + d
a X k^x + d
$\displaystyle y = a \cdot b^{cx} + d$

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
$\displaystyle y = a \cdot b^{-cx} + 12$
(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:
$\displaystyle 62 = a \cdot b^{-c \cdot 0} + 12$

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

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

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

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

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