Solving Log Problems: 2 Quick Challenges

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Homework Help Overview

The discussion revolves around two logarithmic problems presented as challenges by a math teacher. Participants are exploring the properties and definitions of logarithms, particularly in the context of solving equations involving logs. The original poster expresses difficulty with the problems, especially concerning subscripts and the application of logarithmic rules.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants suggest using the definition of logarithms to manipulate the equations. There are attempts to apply properties of logarithms, such as combining logs and transposing terms. Some participants question the correctness of certain steps and assumptions made during the problem-solving process.

Discussion Status

The discussion is active, with participants providing hints and guidance without reaching a consensus on all aspects. Some have successfully solved the first problem, while others continue to explore the second problem, raising questions about the quadratic nature of the resulting equation and the validity of certain solutions.

Contextual Notes

Participants note that the problems are not graded but serve as practice before an upcoming test. There is also mention of potential extraneous solutions in the context of quadratic equations, highlighting the importance of verifying answers against the original equations.

lexpar
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I'm new here. You already knew that. Thank you, you, for clicking on my thread in any case. So, I've got these two log problems. My math teacher gave them out as a challenge. The class is relatively new to logs, and I'm having trouble working with these longer questions (longer than "simplify log5+log3"). I think my main issue is with the sub scripts. I don't really know what to do with them. For the first question, where it a normal algebra question, I'd have it no problem. Those logs get in the way though. We've been given the answers, but need to prove them. Any help would be appreciated.

PS: the problems aren't for marks, they're more of a challenge thing before we do our test.

Homework Statement



1: log25+log2(x-3)=4

The Attempt at a Solution



log25+log2(x-3)=4
log25(x-3)=4
log2(5x-15)=4
?
?
x= 31/5

Homework Statement



log3(x+4)=log77-log3(x+2)

The Attempt at a Solution



log3(x+4)=log77-log3(x+2)
?
?
?
x= -9
x=-5
 
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In your first problem, use the fact that log2a = b <===> a = 2b.
In your second problem, transpose the log expression on the right side over to the left side, and use the fact that log77 = 1. Then you'll have a problem very similar to your first problem.
 
Try to just apply definition of log.

If

log_ab = c

then

b = a^c

Edit: Mark beat me.
 
Welcome to PF!

Hi lexpar! Welcome to PF! :smile:
lexpar said:
I1: log25+log2(x-3)=4
…log25+log2(x-3)=4
log25(x-3)=4
log2(5x-15)=4

If this was ln(5x-15) = 4, you'd just do eboth sides, wouldn't you?

ok, so do 2both sides. :wink:
log3(x+4)=log77-log3(x+2)
…
log3(x+4)=log77-log3(x+2)

Hint: log77 = … ? :smile:
 
Thanks all three of you for the ultra speedy reply :D

Figured out the first:

log25+log2(x-3)=4
log25(x-3)=4
log2(5x-15)=4
log25+log2x-log215=log216
log25+log2x=log231
5x=31
x=31/5! :D

As for the second:

log3(x+4)=log77-log3(x+2)
log3((x+4)(x+2))= 1 (or maybe log33?)

Where to go from here?
 
lexpar said:
As for the second:

log3(x+4)=log77-log3(x+2)
log3((x+4)(x+2))= 1 (or maybe log33?)

Where to go from here?

Same method as for the first one! :wink:
 
lexpar said:
log2(5x-15)=4
log25+log2x-log215=log216
log25+log2x=log231

Unfortunately - twice wrong.

log(a-b) is NOT log(a) - log(b).
 
Borek said:
Unfortunately - twice wrong.

log(a-b) is NOT log(a) - log(b).


In that case, what's the correct process? 31/5 is the correct answer, or so we were told.

EDIT: Put a lot of thought into it. Read your earlier posts again:

log2(5x-15)=4
logab=c then b=ac
therefore
5x-15=24
x=31/5


Better? :D
 
Last edited:
lexpar said:
log25+log2(x-3)=4
log25(x-3)=4
log2(5x-15)=4
log25+log2x-log215=log216

Just write log2(5x-15) = log216 :smile:
 
  • #10
31/5 is a correct answer, but you got it by accident.

lexpar said:
log2(5x-15)=4

Try 5x-15 = ... (just apply the definition, nothing more).
 
  • #11
Borek said:
31/5 is a correct answer, but you got it by accident.



Try 5x-15 = ... (just apply the definition, nothing more).


Thanks, I got it :D Check out my last post, I edited it just for you <3
*ahem*

So still having a bit of trouble with the second question. I get to:

log3((x+4)(x+2)=1
log3(x2+6x+8)=1
apply the definition=
x2+6x+8 = 3

But now I've got 2 x's! Where have I gone wrong??
 
  • #12
You listed two x's as correct answers. This is quadratic function. So everything seems to fit.

Apart from the answers, seems to me like -9 is wrong. But I can be wrong, it is 1 a.m. here.
 
  • #13
lexpar said:
Thanks, I got it :D Check out my last post, I edited it just for you <3
*ahem*

So still having a bit of trouble with the second question. I get to:

log3((x+4)(x+2)=1
log3(x2+6x+8)=1
apply the definition=
x2+6x+8 = 3

But now I've got 2 x's! Where have I gone wrong??

So far, so good. Add -3 to both sides, and solve the quadratic equation. It's an easy one so you can factor it.

Be sure to check both of your answers in the original equation, because one of them is extraneous.
 
  • #14
Thank you all for the responses. You guys really run a quality forum here :D

Can I tempt you guys with one more problem? This one wasn't assigned, I'm just curious. I'm looking for a step by step walkthru here if you guys can give it, since this one flies right over my head.

Question:

logc20c/(logc50-logc5)

Answer: log20c

My progress:

logc20c/(logc50-logc5)
logc20c/(logc10)
...


PS: sorry for the late response, was out playing pool with my dad. It's a very important school night activity :p
 
Last edited:
  • #15
lexpar said:
My progress:

logc20c/(logc50-logc5)
logc20c/(logc10)
...
This is fine, but put an = between the two expressions. The numerator could be written as logc(20) + 1, but I don't see that as an improvement.
 
  • #16
Hi lexpar! :smile:

(just got up :zzz: …)
lexpar said:
Answer: log20c

My progress:

logc20c/(logc50-logc5)
logc20c/(logc10)

Always remember that logca = ln(a)/ln(c) = log(a)/log(c) = logba/logbc for any b. :wink:

In this case, logc20c/(logc10) = … ? :smile:
 
  • #17
...

I'm really not catching on to that one :C
 
  • #18
lexpar said:
...

I'm really not catching on to that one :C

Write each of the top and the bottom of logc20c/(logc10) as fractions.
 

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