Question

How do you solve for x?: `log(x) + log(x-9) = 1`

Answer 1

`x=10`

Explanation:

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

`•color(white)(x)logx+logy=log(xy)`

`•color(white)(x)log_b x=nhArrx=b^n`

`"note that "logx-=log_(10)x`

`rArrlog_(10)x(x-9)=1`

`rArrx^2-9x=10^1=10`

`rArrx^2-9x-10=0larrcolor(blue)"in standard form"`

`rArr(x-10)(x+1)=0`

`rArrx=10" or "x=-1`

`"note that "x>0" and "x-9>0`

`rArrx=-1" is invalid"`

`rArrx=10" is the solution"`