CPhill

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 ^{log_a 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 ^{ln_e √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!!!

We can't take the log of a negative number or zero, so ... x +5 must be > 0

Therefore we have 

x+5 > 0  subtract 5 from both sides

x > -5

So the domain relates to all the x values that are possible in this expression and that would be all "x's" greater than -5   .... we can write this as....

(-5, ∞)

And that's the domain

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.

Note that a log is just an exponent.....for instance, if  I had the following....

log₄ 16    ....this is just asking us...."What power do I need to put on 4 to raise it to 16??"

And the answer is 2 because 4² = 16......so thinking "backwards"...we have

8 ⁴ = 4096      So to write this in log form, we have

Log₈ 4096 = 4    .....Note .....4 is the power that I put on 8 to raise it to 4096

Does that make sense???

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

OK...so we have

2x = 8/9

And we're trying to solve for x.....we've finished the "square root" part

What if I multiplied both sides by 1/2.....what would happen???

1)  log₈ 4096 = 4

2)  8² = 64

We need to take the tangent inverse for this...so we have

atan(12.7) = 85.497810507959°

Note that the tangent is positive in the 3rd quadrant, too.... so this angle might also be

(180 + 85.497810507959)  = 265.497810507959°

Did you get it correct???

Your way would work...the error that you made is when you multiplied both sides by v...you forgot to multiply the "w" by "v,"   too!!

Try it again and see what happens......