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    Hello! New to this so not too sure how to work it was just wondering if any of you clever students out there could help me with this question:
    Solve for x, round your answers to 2 dec places,
    ln(x) + ln(4x) - ln(2x) = 1
    Hope someone can help! Thank you!
    Beth.
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    Use the logarithm laws to combine the LHS into one logarithm and then raise e to the power of both sides of your equation e.g.

    If \mathrm{ln}(x) = 5 then e^{\mathrm{ln}(x)} = e^5 i.e. x=e^5
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    (Original post by BethArkless)
    Hello! New to this so not too sure how to work it was just wondering if any of you clever students out there could help me with this question:
    Solve for x, round your answers to 2 dec places,
    ln(x) + ln(4x) - ln(2x) = 1
    Hope someone can help! Thank you!
    Beth.
    Isn't this the same as the question you asked on the other thread?
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    use the 2 rules that:

    ln(a)+ln(b)=ln(ab)

    and ln(c)-ln(d)=ln(\frac{c}{d})

    (hint: do the plus part first - as you visualise the question as:

    [ln(x)+ln(4x)]-ln(2x)=ln(e)

    ...from there, you shouldn`t need to exponentiate...
 
 
 
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