Question

How do you solve `log_10(8x)-log_10(1+sqrtx)=2`?

Answer 1

`x~~180.3838`

Explanation:

Two thinks to remember here:

[1]`color(white)("XXX")color(red)(log_b p - log_b q = log_b p/q)`

[2]`color(white)("XXX")color(blue)(log_b s = t hArr b^t = s)`

Therefore
`color(white)("XXX")color(red)(log_(10) (8x)-log_(10)(1+sqrt(x)) =2)`
means
`color(white)("XXX")color(red)(log_(10) ((8x)/(1+sqrt(x))) =2)`

and
`color(white)("XXX")color(blue)(log_(10) ((8x)/(1+sqrt(x)))=2)`
means
`color(white)("XXX")color(blue)(10^2=(8x)/(1+sqrt(x))`

From this we can get:
`color(white)("XXX")8x=100(1+sqrt(x))`

`color(white)("XXX")2x=25+25sqrt(x)`

`color(white)("XXX")2x-25sqrt(x)-25=0`

If we (temporarily) let `p=sqrt(x)`
`color(white)("XXX")2p^2-25p-25=0`

then applying the quadratic formula
`color(white)("XXX")p=(25+-sqrt((-25)^2-4 * 2 * (-25)))/(2 * 2)`
and with some simplification
`color(white)("XXX")p=(25+-5sqrt(33))/4`
(note, however the negative version can be eliminated as extraneous since `p=sqrt(x) rarr p >=0` for any `x in RR`)

Therefore
`color(white)("XXX")sqrt(x)=(25+5sqrt(33))/4`

and
`color(white)("XXX")x=((25+5sqrt(33))/4)^2 ~~180.3838`
(using a calculator for this last step)