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Jadedry
 one year ago
Logarithm and Indices Question!
Jadedry
 one year ago
Logarithm and Indices Question!

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Jadedry
 one year ago
Best ResponseYou've already chosen the best response.1Solve for x, correct to two significant figure: \[4^{2X+1} * 5^{X2} =6 ^{1X}\] 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!

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.3\[4^{2x+1}=\frac{ 6^{1x} \times 5}{ 5^{x1} }=>4^{2x+1}=6^{1x}\times 5^{1x} \times 5\] can u do it from here :)

Jadedry
 one year ago
Best ResponseYou've already chosen the best response.1I tried it again but I'm still stuck, sorry. =/ What is the next step?

Jadedry
 one year ago
Best ResponseYou've already chosen the best response.1Wait, 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?

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.3hey forget that^ thing cause it ws getting too messy :/ and do it like this \[\log_{2}(4^{2x+1}\times5^{x2})=\log_{2}6^{1x}\]\[\log_{2}4^{2x+1} +\log_{2}5^{x2}=(1x)\log_{2}2 +(1x)\log_{2}3\]\[4x+2+(x2)\log_{2}5=1x+\log_{2}3xlog_{2}3\]\[4x+2+xlog_{2}52\log_{2}5=1x+\log_{2}3xlog_{2}3\]\[x(4+\log_{2}5+1+\log_{2}3)=2\log_{2}52+1+\log_{2}3\]

Jadedry
 one year ago
Best ResponseYou've already chosen the best response.1So:\[4^{1x} * 4^{3x} = 6^{1x} * 5^{1x} * 5\] \[\frac{ 4^{1x} }{ 6^{1x } * 5^{1x } } = \frac{ 5 }{ 4^{3x} }\]

Jadedry
 one year ago
Best ResponseYou've already chosen the best response.1Alright, this is making a bit more sense, so you actually multiplied one of the logs on the right side by (1x) 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:

Jadedry
 one year ago
Best ResponseYou've already chosen the best response.1So \[\frac{ \log_{2}75  1 }{ \log_{2} 15 + 5 } = x\] = 0.587 = 0.59 Yes! Correct answer, thanks for your time and trouble @imqwerty ! ;U;
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