[MOVED] solve log eqn: log_16 (6x - 5) + log_16 (x) = 1/2

Jade

Junior Member
Joined
Sep 16, 2006
Messages
95
Solve the logarithmic equation:

. . .log_16 (6x - 5) + log_16 (x) = 1/2

I am having some issue with solving this. My thoughts are that I should rewrite it and solve.
 
Where's your trouble?.

\(\displaystyle \L\\log_{16}(6x-5)+log_{16}(x)=\frac{1}{2}\)

Rule of logs(multiplication and addition relation):

\(\displaystyle \L\\log_{16}(x(6x-5))=\frac{1}{2}\)

Base 16:

\(\displaystyle \L\\x(6x-5)=16^{\frac{1}{2}}=4\)

Resulting quadratic:

\(\displaystyle \L\\6x^{2}-5x-4=0\)

Solve away. You'll have 2 solutions. One negative, one positive. One will be extraneous. Know which?.
 
Factor out

(3x-4)(2x+1)
x=4/3 x=-1/2

I think I am having trouble figuring out the log rules. I have a copy of what the log rules should be in front of me...

logb 1 = 0
logb b = 1
logb (xy) = logb x + logb y
logb (x/y) = logb x - logb y
logbx^r = r logb x

Then I have the special cancel out rule logb b^r = r logb b you would cancel out the b and be left with just r

I think I am having trouble putting these rules in action.

Same when e is involved.... for example I have 1+2e^5x=15
so this is what I want to do...
2e^5x=14
e^5x=7
lne^5x=ln7
x=ln7/5
x=.39

Am I getting the hang of this???
 
Jade said:
(3x-4)(2x+1)
x=4/3 x=-1/2
What happened to the "=0" part of the equation?

You were warned that one of these solutions would not be valid. Which one? Do you understand why?

Jade said:
I think I am having trouble figuring out the log rules.
The rules aren't meant to be anything "deep". Just memorize them.

Jade said:
logb 1 = 0
logb b = 1
Assuming you mean "logb" to be "log<sub>b</sub>", these "rules" come from the basic definition of logarithms. Since b<sup>0</sup> = 1, then log<sub>b</sub>(1) = 0; since b<sup>1</sup> = b, then log<sub>b</sub>(b) = 1.

Jade said:
logb (xy) = logb x + logb y
logb (x/y) = logb x - logb y
logbx^r = r logb x
"Times" inside is "plus" outside; "divide" inside is "minus" outside; "to the power" inside is "times" outside. Just memorize 'em.

Jade said:
Same when e is involved.... for example I have 1+2e^5x=15
What you have posted means "1 + 2e<sup>5</sup>x = 15". Is that what you meant?

Eliz.

(In the future, please post new questions as new threads, not as replies to old threads. Thank you.
 
What I have posted

No - let me try to post the original again

1 + 2e(^5x) = 15

As far as the log rules - I will try to imprint them on my brain :D
 
Jade said:
1 + 2e(^5x) = 15
Do you mean "1 + 2e^(5x)" (that is, 1 + 2e<sup>5x</sup>)?

Assuming so, then your solution method appears to be correct. But the answer should probably be left in "exact" form, rather than being replaced with a two-decimal-place approximation.

Eliz.
 
Top