Question

Why is it that -log(x+8)=4-log(x-7) has no solution? (they are log base 2)

Answer 1

Note that -log(x+8) + log(x-7) = 4, and that the left side is equal to

x-7
log -------------
x-8

Therefore,

x-7
log ------------- = 4
x-8

Acknowledging that your "log" actually represents "log to the base 2 of ... "

We get:

x-7
--------- = 2^4 = 16
x-8

Can this be solved for x?

Rearranging, x-7 = 16x - 128, or -7 = 15x - 128, or 121 = 15x
121
Dividing 121 by 15, we get x = ------- = 121/15 = approx. 8.067.
15


So far I see no reason why the given -log(x+8)=4-log(x-7) "has no solution."

Answer 2

`-log(x+8)=4-log(x-7)`


`-\log _(10)\left(x+8\right)+\log _(10)\left(x+8\right)=4-\log _(10)\left(x-7\right)+\log _(10)\left(x+8\right)`


`0=4-\log _(10)\left(x-7\right)+\log _(10)\left(x+8\right)`


`0+\log _(10)\left(x-7\right)=4-\log _(10)\left(x-7\right)+\log _(10)\left(x+8\right)+\log _(10)\left(x-7\right)`


`\log _(10)\left(x-7\right)=\log _(10)\left(x+8\right)+4`


`\log _(10)\left(x-7\right)=\log _(10)\left(x+8\right)+\log _(10)\left(10000\right)`


`x-7=\left(x+8\right)\cdot \:10000`


`\mathrm{Solve\:}\:x-7=\left(x+8\right)10000:\quad x=-(26669)/(3333)`


`\mathrm{Verifying\:Solutions}:\quad x=-(26669)/(3333)\space\mathrm{False}`


`\mathrm{No\:Solution\:for\:x\in \mathbb{R}}`