how would you solve: log x + log (x-9) = 1?

Answer 1

Exponentiating both sides might work, but 10^(x+y) does not equal 10^x + 10^y. It equals 10^x * 10^y.

10^(log(x)+log(x-9)) = 10^1

10^(log(x)) * (10^(log(x-9))) = 10

x * (x-9) = 10

x^2 - 9x -10 = 0

x = 10, x = -1

Only x=10 gives non-negative values for the log fcn.

To check it:

log(10) + log(10-9) = 1

1 + 0 = 1

Answer 2

Remember that for logs of the same base, log(ab) = log(a) + log(b), so this is

log( x(x-9) ) = 1
x(x-9) = 10^1
x^2 - 9x - 10 = 0
(x - 10)(x + 1) = 0
so x = 10, -1

Somebody suggested doing an antilog of both sides by taking 10^(log x) + 10^(log (x-9)) = 10^1, but you can't distribute exponents like this on the left. For example, 1+1=2, but it's not true that 10^1 + 10^1 = 10^2

Answer 3

Take logs as being to base ten.
log [ x.(x - 9) ] = 1
x.(x - 9) = 10^1
x² - 9x - 10 = 0
(x - 10).(x + 1) = 0
x = - 1 , x = 10

Answer 4

have both sides as exponents of 10

10^log x + 10^ log(x-9) = 10^1

2x-9=10
2x=19
x=19/2