Solving for x: log(3x - 7) + log(x - 2) = 1

[Bladesofhalo]

Ok, so I have a simple log equation:
log(3x - 7) + log(x - 2) = 1

Now using properties of logs, I get:
log(3x[sup:2gwuj0vm]2[/sup:2gwuj0vm] + x - 14) = 1

Now how would I proceed with this? do I have to assume log[sub:2gwuj0vm]10[/sub:2gwuj0vm]1 = 0 in order to solve?

 

[Loren]

log 10 = 1???

 

[stapel]

do I have to assume log[sub:2l1g3iyz]10[/sub:2l1g3iyz]1 = 0 in order to solve?

At this point, yes, you're going to have to assume the base to be some value. (Since the right-hand side is "equals 1", I'm not sure where you're getting the zero from...?)

If, in your particular text, a log with no stated base is assumed to have base ten, then, using the fact that 10[sup:2l1g3iyz]1[/sup:2l1g3iyz] = 10 (so log[sub:2l1g3iyz]10[/sub:2l1g3iyz](10) = 1), you would have the following:

. . . . .log[sub:2l1g3iyz]10[/sub:2l1g3iyz][(3x - 7)(x - 2)] = log[sub:2l1g3iyz]10[/sub:2l1g3iyz](10)

Then, setting the arguments equal, you would have:

. . . . .(3x - 7)(x - 2) = 10

. . . . .3x[sup:2l1g3iyz]2[/sup:2l1g3iyz] - 13x + 14 = 10

...and so forth.

Eliz.

 

[Bladesofhalo]

Thank you eliz and loren, I appreciate the help very much.