What is the solution of log(x-21) = 2 - logx

First we will find which values x should not exist.

Log(x -21)

Well,

x - 21 > 0

Then,

x > 21


Now, we let's to the question

First,

Rewrite 2 = Log 100

Then,

Log (x -21) = Log 100 - Log(x)

Use the property:

Log(a) - Log(b) = Log( a / b)

Then,

Log( 100) - Log( x) = Log(100/x)

So us wiil stay:

Log(x - 21) = Log(100/x)

Now cancel the Logs...

x - 21 = 100/x

Isolating x to the letf side:

x( x - 21) = 100

Applying the distributive on the member firft:

x^2 - 21x = 100

Passing 100 to the left side of equation us will have:

x^2 - 21x - 100 = 0

Now we've a equation of baskara:

Where,

a = 1
b = -21
c = -100

So follow :

Delta = b^2 -4ac

Delta = (-21)^2 -4.(1).(-100)

Delta = 441 + 400

Delta = 841

As x = [ -b +/- √( Delta) ] /2a

Then,

x = [ -(-21) +/- √(841) ]/2.(1)

x = [ 21 +/- 29 ]/2



1:

x = (21 - 29)/2

x = -4

Or

x = (21 + 29)/2

x = 25


As X > 21

Then,

x = - 4 not exist.

Just x = 25 would be the answer.

(-0, -7) is your answer