## log_6 (x+3) + log_6 (x+4) = 1, for X

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### log_6 (x+3) + log_6 (x+4) = 1, for X

Solve log_6 (x+3) + log_6 (x+4) = 1, for X

Okay! I know the answer is -1, this is for my review. My issue is I don't know WHY the answer is -1. I can't work the problem to get that answer.

I move the second logarithm to the right side of the equation, getting log_6(x+3) = -log_6(x+4) + 1

And because the bases are the same, I set x+3 = -(x+4) + 1, but this works out to x=-3, which isn't the answer.

What am I doing wrong?
Wolvenmoon

Posts: 2
Joined: Tue May 05, 2009 2:45 am

### Re: log_6 (x+3) + log_6 (x+4) = 1, for X

It figures, 5 minutes after I post I get it.

The problem is that I was ignoring another property of logarithms. log_6(x+3) + log_6(x+4) = 1 is the same as log_6(x+3)(x+4) = 1. To find X, I set (x+3)(x+4) = 6, and solve. This factors down to (x+1)(x+6), which is x=-1, x=-6. Because you cannot take a negative logarithm, x=-6 is thrown out as it breaks BOTH logarithms, so the answer is -1.

Wolvenmoon

Posts: 2
Joined: Tue May 05, 2009 2:45 am

Wolvenmoon wrote:Solve log_6 (x+3) + log_6 (x+4) = 1, for X

Use log rules to compress the left-hand side:

. . . . .$\log_6(x\, +\, 3)\, +\, \log_6(x\, +\, 4)\, =\, \log_6\left((x\, +\, 3)(x\, +\, 4)\right)$

Then use the definition of logs to convert the log equation into the equivalent exponential equation:

. . . . .$(x\, +\, 3)(x\, +\, 4)\, =\, 6^1$

. . . . .$x^2\, +\, 7x\, +\, 12\, =\, 6$

. . . . .$x^2\, +\, 7x\, +\, 6\, =\, 0$

Then factor the quadratic to get:

. . . . .$(x\, +\, 6)(x\, +\, 1)\, =\, 0$

This tells you that x = -6 or else x = -1. But since the arguments of the logs are "x + 3" and "x + 4", then "x = -6" would give you negatives inside the logs, which won't work.

Good job! (And thank you for showing your steps and reasoning so nicely!)

stapel_eliz

Posts: 1803
Joined: Mon Dec 08, 2008 4:22 pm