Question

1/log(x+1/x+4) base4≤ 1/log (x+3) base 4 find set of x​

Answer 1

Answer:

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list question solution is here

log(x-1)base4 = 1/2 log(x-1)(base2)

therefore

1/2log(x-1)base2 = log(x-3)(base2)

therefore

log(x-1)base2 = 2log(x-3)base2 = log(x-3)^2(base2)

since both sides have the same base therefore we can remove the log

we get

(x-1)=(x-3)^2 = x^2 - 6x +9

on solving further we get

x^2-7x+10=0

x^2-2x-5x+10=0

x(x-2) -5(x-2)=0

(x-5)(x-2)=0

or x=5 or x=2

iind solution is here

logx(base8)+log(x-4)base8+log(x-6)base8=2

then

log x(x-4)(x-6)(base8)=2

or

x(x-4)(x-6)=64

on solving we get a cubic equation

x^3 -10x^2 +24x - 64=0

let the roots of the given eq. be a,b and c

then a + b+c= 10

& abc = 64

now by a.m.-g.m. inequality we get

(a+b+c)> Or = 3(abc)^1/3 = 3*4 = 12

which is less than a+b+c

hence no solutions are possible

i hope you have understood the solutions

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