Question

If `log_ba=1/x and log_a √b =3x^2`, show that x =1/6?

Answer 1

`"see explanation"`

Explanation:

`"using the "color(blue)"law of logarithms"`

`•color(white)(x)log_b x=nhArrx=b^n`

`log_b a=1/xrArra=b^(1/x)`

`log_a b^(1/2)=3x^2rArrb^(1/2)=a^(3x^2)`

`"substitute "a=b^(1/x)" into "a^(3x^2)`

`b^(1/2)=(b^(1/x))^(3x^2)=b^(3x)`

`"we have "b^(1/2)=b^(3x)`

`3x=1/2`

`"divide both sides by 3"`

`x=(1/2)/3=1/6`

Answer 2

`x=1/6`

Explanation:

`log_b a=1/x`

So

`xlog_ba=1`

`x=1/log_b a`

`x=log_ab`

Now

`1/2log_a b=3x^2" "` (substituting from `log_a b=x`)

So

`x/2=3x^2`

`x=6x^2`

`6x=1`

`x=1/6` with `x !=0`

Hence proved.

Answer 3

We will start the proof

Explanation:

We have
`ln(a)/ln(b)=1/x`
and

`1/2*ln(b)/ln(a)=3x^2`
so

`ln(a)/ln(b)=1/(6x^2)`

using the first equation we get
`1/x=1/(6x^2)`

multiplying by `x^2` we get

`x=1/6` for `xne 0`