Solving Log Problems: 2 Quick Challenges

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In summary, 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.
  • #1
lexpar
11
0
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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  • #2
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.
 
  • #3
Try to just apply definition of log.

If

[tex]log_ab = c[/tex]

then

[tex]b = a^c[/tex]

Edit: Mark beat me.
 
  • #4
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:
 
  • #5
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?
 
  • #6
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:
 
  • #7
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).
 
  • #8
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:
  • #9
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.
 

What are logarithms and why are they useful?

Logarithms are mathematical functions that can be used to solve exponential equations. They are useful because they allow us to simplify complex calculations, such as multiplying and dividing large numbers, into more manageable equations.

What is the difference between natural logarithms and common logarithms?

Natural logarithms use the base e (approximately 2.718) and are denoted as ln(x). Common logarithms use the base 10 and are denoted as log(x). They both have their own unique properties and can be used for different types of calculations.

How do I solve a logarithmic equation?

To solve a logarithmic equation, you can use the properties of logarithms to rewrite the equation in a simpler form. Then, you can use algebraic techniques to isolate the variable and solve for its value. Remember to check your solution to ensure it is valid.

What are some real-world applications of logarithms?

Logarithms have many real-world applications, including population growth and decay, measuring the intensity of sound and earthquakes, and calculating pH levels. They are also used in financial calculations, such as compound interest and the decibel scale.

What are common mistakes to avoid when solving logarithmic equations?

Some common mistakes to avoid when solving logarithmic equations include forgetting to check for extraneous solutions, not using the properties of logarithms correctly, and making errors when simplifying the equation. It is important to carefully follow the steps and double-check your work to avoid these mistakes.

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