Evaluate the expression.

log_{9} √9

log_{5} 0.2

5^{log5 19}

e^{ln }

11 |

ln(1/e)

sally1 Jul 9, 2014

#5**+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) log_{5} 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.......

log_{5} 5^{-1} and by a log property, we can bring the -1 out front ..... and we have.....

(-1)log_{5} 5 ....and log_{5} 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 log_{2} 4 just asks..what power do we need to raise 2 to to get 4??? The answer is 2

So we really have

2^{2} = 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

#1**+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

CPhill Jul 9, 2014

#2**0 **

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

sally1 Jul 9, 2014

#3**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.

CPhill Jul 9, 2014

#5**+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) log_{5} 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.......

log_{5} 5^{-1} and by a log property, we can bring the -1 out front ..... and we have.....

(-1)log_{5} 5 ....and log_{5} 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 log_{2} 4 just asks..what power do we need to raise 2 to to get 4??? The answer is 2

So we really have

2^{2} = 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