Calculus: Solve log(22⁢x+6/2⁢x+24)= log(10) for x?

Question

Can someone give me the properties I need to solve this equation? I do not want the actual answer. Thanks!

Please note that I stated I did NOT want the value of x!!!

Answer 1

I think you meant the thing inside the log to be (22x+6)/(2x+24), not 22x + 6/2x + 24.
If so, then all you need to know is that log a = log b is just the same as a = b. So you can eliminate the logs immediately, and then just solve for x (get rid of fractions, put all xs on one side, and then solve).

Answer 2

All you have to do is take the antilog of both sides, being left with:

22⁢x+6/2⁢x+24= 10

Move everything to the left hand side:

22⁢x+6/2⁢x+14 = 0

And then solve as normal. Once you get a value for x, be sure to test the value in the original equation to see if you end up taking the log of a negative number. If that's the case, discard it.

Answer 3

Take anti-log on both sides and solve for x. That is:
22x+6/2x+24 = 10

If you need some log properties- have a look
http://people.hofstra.edu/faculty/Stefan_Waner/RealWorld/calctopic1/logs.html

Answer 4

If log(a) = log(b), then a=b.
Therefore,
(22x+6)/(2x+24) = 10
Multiply both sides by 2x+24
22x + 6 = 10(2x+24)
22x + 6 = 20x + 240
Subtract 20 from each side
2x + 6 = 240
Subtract 6 from each side
2x = 234
Divide both sides by 2
x = 117