How do I solve log(x-3)=1-log(x)?

Answer 1

log(x-3)=1-log(x)
log(x-3) + log(x) =1
log((x-3)x) =1
(x-3)x = 10
x^2 -3x -10 =0
(x-5)(x+2) = 0
=> x = 5 or x = -2,
but x= -2 is not a solution of the original problem, since log (-2) is NOT defined, as well as log (-2-3)!.
so the only solution is x=5. .

Answer 2

To solve any equation involving logs, you need to keep in mind these three log rules:

(1) log[base b](ac) = log[base b](a) + log[base b](c)

Translation: The log of a product is equal to the sum of the logs of the factors.

(2) log[base b](a/c) = log[base b](a) - log[base b])(c)

Translation: The log of a quotient is equal to the difference of the logs of the numerator and denominator (for c nonzero).

(3) log[base b](a^c) = c * log[base b](a)

Whenever you have a power inside of a log, you can effectively take that power out and multiply it outside of the log.

Remember that these log rules work both ways; you can go from one to the other at any time, so long as you meet the conditions.

You should also know how to change from exponential to logarithmic form.

log[base b](a) = c

Is the same as

b^c = a

The thing to note when doing the conversion is THE BASE OF THE LOG BECOMES THE BASE OF THE EXPONENT. Everything else you can just memorize.

Now, onto your problem.

log(x - 3) = 1 - log(x)

Your first step is to bring all logs to the left hand side.

log(x - 3) + log(x) = 1

Using one of the above log rules, combine into a single log.

log[ (x - 3)x ] = 1

And now change into exponential form. Our log is base 10, since it wasn't given.

10^1 = (x - 3)x

Solve the left hand side, expand the right hand side,

10 = x^2 - 3x

Bring the 10 to the right hand side,

0 = x^2 - 3x - 10

And now, factor.
0 = (x - 5)(x + 2)

This means x = 5 or x = -2
BUT WAIT! We CANNOT assume both of these values will work, because we CANNOT take the log of a negative number. What we would do is TEST these two values into the original question; if it results in taking the log of a negative number, we discard that solution.

Test x = 5:

log (5 - 3) = 1 - log(5)

This checks out.

Test x = -2. We can immediately see that this one fails (on the right hand side, we have 1 - log(x), and if we were to plug in -2 for that log, it would immediately be invalidated).

Therefore, our only solution is x = 5.

Answer 3

I'll use x2 to represent x squared:

log(x-3) = 1 - log(x)
log(x-3) + log(x) = 1
log(x2 - 3x) = 1 [addition combines and becomes multiplication]
x2 - 3x = 10 [converting to exponential form]
x2 - 3x - 10 = 0
(x-5)(x+2) = 0 [factoring]
x = 5 or x = -2
but, x = -2 is not a solution since you cannot take the log of a negative number.
so, x = 5

Answer 4

log (x-3) = 1-log x
log (x-3) + log (x) = 1
log [(x-3)(x)] = 1
log (x^2 - 3x) = 1
10^1 = x^2 - 3x
so
x^2 - 3x = 10
x (x-3) =10
thus
x=10 or 13

Answer 5

Hint: 1 = log(10)

So you have:

log(x-3) = log(10)-log(x)

Answer 6

log(x-3)+log(x)=1

log(x-3(x))=1

log(x^2-3x)=1

(base of log)^1=x^2-3x

0=x^2-3x-(base of log)^1
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Note I do not know what your base of log is but normally they don't give it when it is the standard 10, but say you mean 10 then solve the quadratic

x^2-3x-10=0

(x-5)(x+2)=0

x=5 only solution because
you can't have negative log when x=-2