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Math Help - (Aust)Yr11 logs assignment (Due nxt Thur)

  1. #1
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    (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!
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  2. #2
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    Hello, mathsdunce000!


    \begin{array}{cc}\text{Eq. 1} & 5\!\cdot\!2^{x+1} - 12 \:= \:0\\<br />
\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: . 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



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

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  3. #3
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    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!
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  4. #4
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    Quote Originally Posted by mathsdunce000 View Post
    eq2) 3x2^x = 4x3^(-x)+7
    Quote Originally Posted by mathsdunce000 View Post
    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.
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    thank you soroban and squark for your help! Any ideas on the graph?
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  6. #6
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by mathsdunce000 View Post
    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
    Your form is
    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
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