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Evaluate the expression.

  log9 √9

 log5 0.2

5log5 19

eln 

11

 

ln(1/e)

 

 Jul 9, 2014

Best Answer 

 #5
avatar+124676 
+5

The point here, sally1, is that we don't need a calculator for ANY of them....!!! We're just using some log properties to evaluate them.....here's the second one....

2) log5 0.2    ..... this one looks tricky, but let's remember something...that 0.2 = 2/10 = 1/5

And we can write 1/5   as  5-1   (remember that ??)

So we have.......

log5 5-1  and by a  log property, we can bring the -1 out front  ..... and we have.....

(-1)log5 5    ....and log5 5  = 1  ...so we have....

(-1)(1)  = -1   

3) These used to give me trouble...once you see the 'trick,' it's easy......let me give you an example...let's say we have 

2 log2 4          

Note that log2 4   just asks..what power do we need to raise 2 to to get 4???  The answer is 2

So we really have

22  = 4      Note that this was just the last number in the original exponent!!!

In general   when we have  .....a loga b  the answer is just "b".....this is one you may have to think about a little bit!!!

So.....in your problem  5 log5 19    ...the answer is just "19"

4) e ln √11   Note that there is a little "e" implied here  we can write this as

e lne √11      And we have the same situation as the previous problem!!!

The answer is just "√11"

 

Lastly

5)  ln (1/e)  = ln (e) -1     and by a property of exponents we can bring the -1 to the front

So we have

(-1) ln e        and ... ln e  = 1  ...   so we have

(-1)(1) = -1

 

And that's it!!!

 

  

 

 

 

      

 

 Jul 9, 2014
 #1
avatar+124676 
+5

1)  Note that √9 = 9 1/2    ....so we have...

Log 9 9 1/2 .... and by a property of logs, we can bring the 1/2 in front of the expression..so...

1/2 Log 9 9 .....and  Log 9 9  = 1  ....so we have

1/2 * 1  = 1/2

 

  

 Jul 9, 2014
 #2
avatar+253 
0

For the log problems I'm trying to put it into the calculator but the answers are not coming out right. 

 Jul 9, 2014
 #3
avatar+124676 
0

That's probably because you're trying to use base 10......most of these (like the first one, for instance) are using a different base.

 

  

 Jul 9, 2014
 #4
avatar+253 
0

So how would I be able to evaluate it?

 Jul 9, 2014
 #5
avatar+124676 
+5
Best Answer

The point here, sally1, is that we don't need a calculator for ANY of them....!!! We're just using some log properties to evaluate them.....here's the second one....

2) log5 0.2    ..... this one looks tricky, but let's remember something...that 0.2 = 2/10 = 1/5

And we can write 1/5   as  5-1   (remember that ??)

So we have.......

log5 5-1  and by a  log property, we can bring the -1 out front  ..... and we have.....

(-1)log5 5    ....and log5 5  = 1  ...so we have....

(-1)(1)  = -1   

3) These used to give me trouble...once you see the 'trick,' it's easy......let me give you an example...let's say we have 

2 log2 4          

Note that log2 4   just asks..what power do we need to raise 2 to to get 4???  The answer is 2

So we really have

22  = 4      Note that this was just the last number in the original exponent!!!

In general   when we have  .....a loga b  the answer is just "b".....this is one you may have to think about a little bit!!!

So.....in your problem  5 log5 19    ...the answer is just "19"

4) e ln √11   Note that there is a little "e" implied here  we can write this as

e lne √11      And we have the same situation as the previous problem!!!

The answer is just "√11"

 

Lastly

5)  ln (1/e)  = ln (e) -1     and by a property of exponents we can bring the -1 to the front

So we have

(-1) ln e        and ... ln e  = 1  ...   so we have

(-1)(1) = -1

 

And that's it!!!

 

  

 

 

 

      

 

CPhill Jul 9, 2014
 #6
avatar+253 
0

Thank you so much. Now, I'm about to go and try to work some out on my own. Thanks for your help.

 Jul 9, 2014

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