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C2 - Logarithms

Hi, I'm stuck on these logarithm questions and hope someone can help me

1)

a) If logx²-logxy+log5x²=0, find an equation involving log x and log y

b) Hence find y in terms of x.

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2) The variable x satisfies the equation (3^x).[4^(2x+1)]=6^(x+2)

By taking logarithms of both sides show that x= log9/log8

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Given that 2In(2x+2)+Inx=In(13x+6) prove that x³+4x²-9x-6=0. Hence solve the eqation 2In(2x+2)+Inx=In(13x+6). (I have already found the roots of the x³+4x²-9x-6=0 equation, which are 2, -3 (+-) √6)

Thank you very much for your help! :smile:
sweet_gurl
Hi, I'm stuck on these logarithm questions and hope someone can help me

1)

a) If logx²-logxy+log5x²=0, find an equation involving log x and log y

b) Hence find y in terms of x.

-----------------------------------

2) The variable x satisfies the equation (3^x).[4^(2x+1)]=6^(x+2)

By taking logarithms of both sides show that x= log9/log8

-----------------------------------

Given that 2In(2x+2)+Inx=In(13x+6) prove that x³+4x²-9x-6=0. Hence solve the eqation 2In(2x+2)+Inx=In(13x+6). (I have already found the roots of the x³+4x²-9x-6=0 equation, which are 2, -3 (+-) √6)

Thank you very much for your help! :smile:



a)logx²-logxy+log5x²=0

2log x - log x - log y + 2log x + log 5=0
3 log x - log y + log 5 =0

b)log y = log x + log 5
log y = log 5x
y=5x


2) (3^x).[4^(2x+1)]=6^(x+2)

log [3x (42x+1)]= log 6x+2

x log 3 + (2x+1) log 4 = (x+2) log 6

x log 3 + 2x log 4 - x log 6= 2 log 6 - log 4

log (3x 42x / 6x )= log (6²/4)

log (3x 24x / 3x 2x )= log 9

log (23x )= log 9

x log 23 = x log 8 = log 9

x= log 9/log 8

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2 ln(2x+2)+lnx=ln(13x+6)
ln [(2x+2)²x]=ln(13x+6)
(2x+2)²x=13x+6
4x³+8x²+4x=13x+6
4x³+8x²-9x-6=0 ???
Reply 2
sweet_gurl

2) The variable x satisfies the equation (3^x).[4^(2x+1)]=6^(x+2)

By taking logarithms of both sides show that x= log9/log8

log(3^x).[4^(2x+1)]=log6^(x+2)
xlog3 + (2x+1)log4 = (x+2)log6
xlog3 + 2xlog4 - xlog6 = 2log6 - log4
x(log3 + log16 - log6) = log36 - log4
xlog(3*16/6) = log(36/4)
x = log8/log9