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Thread: Logarithm questions

  1. #1
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    Logarithm questions

    Solve the following logarithmic equations:
    a) log2 (6x-8) - log2 (2x+4) = log2 (x-6)
    b) log3 (x+3) - log3 (x-3) = 1
    c) logn 1 = x

    kinda stuck on those ones
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by kevinz View Post
    Solve the following logarithmic equations:
    a) log2 (6x-8) - log2 (2x+4) = log2 (x-6)
    $\displaystyle \log_2 (6x - 8) - \log_2 (2x + 4) = \log_2 (x - 6)$

    $\displaystyle \Rightarrow \log_2 \left( \frac {6x - 8}{2x + 4} \right) = \log_2 (x - 6)$ ............since $\displaystyle \log_a x - \log_a y = \log_a \left( \frac {x}{y}\right)$

    Now we can just drop the logs:

    $\displaystyle \Rightarrow \frac {6x - 8}{2x + 4} = x - 6$

    I think you can take it from here


    b) log3 (x+3) - log3 (x-3) = 1
    $\displaystyle \log_3 (x + 3) - \log_3 (x - 3) = 1$

    $\displaystyle \Rightarrow \log_3 \left( \frac {x + 3}{x - 3} \right) = 1$

    $\displaystyle \Rightarrow 3^1 = \frac {x + 3}{x - 3}$ ...........since if $\displaystyle \log_a b = c$, then $\displaystyle a^c = b$

    ...and i think you can solve that equation


    c) logn 1 = x
    $\displaystyle x = 0$ right off the bat!

    The $\displaystyle \log_a 1 = 0$ always, i don't care what the base is, the log of 1 is 0, why?

    $\displaystyle \log_n 1 = x$

    $\displaystyle \Rightarrow n^x = 1$

    Anything raised to the zero gives one, so the power is zero
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  3. #3
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    Hello, kevinz!

    Solve the following logarithmic equation:

    $\displaystyle a)\;\log_2(6x-8) - \log_2(2x+4) \:=\:\log_2 (x-6)$

    Here's a trick I've taught all my students . . .

    If we have a logarithmic equation and a log-term is negative,
    . . "move" it to the other side. .**

    So we have: .$\displaystyle \log_2(6x-8) \;=\;\log_2(x-6) + \log_2(2x+4) \;=\;\log_2\left[(x-6)(2x+4)\right]$

    . . . . .Then: .$\displaystyle \log_2(6x-8) \;=\;\log_2(2x^2-8x-24)$


    Exponentiate (or "un-log") both sides: .$\displaystyle 6x - 8 \;=\;2x^2 - 8x - 24$

    We have a quadratic: . . $\displaystyle x^2 - 7x - 8 \;=\;0$

    . . . . . which factors: .$\displaystyle (x - 8)(x + 1) \;=\;0$

    . . . . .and has roots: . . . . $\displaystyle x \:=\:8,\:(-1)$ .
    -1 is extraneous

    Therefore, the only root is: .$\displaystyle x \,=\,8$


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    ** .Why? .To avoid that annoying quotient.

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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Soroban View Post
    Hello, kevinz!


    Here's a trick I've taught all my students . . .

    If we have a logarithmic equation and a log-term is negative,
    . . "move" it to the other side. .**

    So we have: .$\displaystyle \log_2(6x-8) \;=\;\log_2(x-6) + \log_2(2x+4) \;=\;\log_2\left[(x-6)(2x+4)\right]$

    . . . . .Then: .$\displaystyle \log_2(6x-8) \;=\;\log_2(2x^2-8x-24)$


    Exponentiate (or "un-log") both sides: .$\displaystyle 6x - 8 \;=\;2x^2 - 8x - 24$

    We have a quadratic: . . $\displaystyle x^2 - 7x - 8 \;=\;0$

    . . . . . which factors: .$\displaystyle (x - 8)(x + 1) \;=\;0$

    . . . . .and has roots: . . . . $\displaystyle x \:=\:8,\-1)$ .
    -1 is extraneous

    Therefore, the only root is: .$\displaystyle x \,=\,8$


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    ** .Why? .To avoid that annoying quotient.

    Quotients are annoying, aren't they?! didn't seem like too big a deal in this case though, these were simple enough
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