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Jadedry

  • one year ago

Logarithm and Indices Question!

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  1. Jadedry
    • one year ago
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    Solve for x, correct to two significant figure: \[4^{2X+1} * 5^{X-2} =6 ^{1-X}\] I think I'll need to use a logarithm in order to sort it out, but I;m confused as to how to start, does anyone have a few hints? I'm more interested in the method than the answer. c; Thanks in advance!

  2. imqwerty
    • one year ago
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    \[4^{2x+1}=\frac{ 6^{1-x} \times 5}{ 5^{x-1} }=>4^{2x+1}=6^{1-x}\times 5^{1-x} \times 5\] can u do it from here :)

  3. Jadedry
    • one year ago
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    I tried it again but I'm still stuck, sorry. =/ What is the next step?

  4. Jadedry
    • one year ago
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    Wait, so I have to divide \[4^{2x+1}\]by \[4^{3x}\] And then divide the right hand side by the same number and /then/ deal with the logarithms?

  5. imqwerty
    • one year ago
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    hey forget that^ thing cause it ws getting too messy :/ and do it like this- \[\log_{2}(4^{2x+1}\times5^{x-2})=\log_{2}6^{1-x}\]\[\log_{2}4^{2x+1} +\log_{2}5^{x-2}=(1-x)\log_{2}2 +(1-x)\log_{2}3\]\[4x+2+(x-2)\log_{2}5=1-x+\log_{2}3-xlog_{2}3\]\[4x+2+xlog_{2}5-2\log_{2}5=1-x+\log_{2}3-xlog_{2}3\]\[x(4+\log_{2}5+1+\log_{2}3)=2\log_{2}5-2+1+\log_{2}3\]

  6. Jadedry
    • one year ago
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    So:\[4^{1-x} * 4^{3x} = 6^{1-x} * 5^{1-x} * 5\] \[\frac{ 4^{1-x} }{ 6^{1-x } * 5^{1-x } } = \frac{ 5 }{ 4^{3x} }\]

  7. Jadedry
    • one year ago
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    Alright, this is making a bit more sense, so you actually multiplied one of the logs on the right side by (1-x) in order to "detach" it? Neat trick. Give me a few minutes, I think I should be able to get it but let me make doubly sure. c:

  8. Jadedry
    • one year ago
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    So \[\frac{ \log_{2}75 - 1 }{ \log_{2} 15 + 5 } = x\] = 0.587 = 0.59 Yes! Correct answer, thanks for your time and trouble @imqwerty ! ;U;

  9. imqwerty
    • one year ago
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    (:

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