1. ## exponential

I'm having a horrible problem with logarithms...rather natural logarithms. Okay so here's the problem.

5(4^(2x-1))=21

Now I have the soloution, but I need to understand how they got to the solution. There is one other problem, and I'm not sure about it....

But that's a posting for after we work on this one.

2. ## dont scream at me if i got it wrong or haven't answer ur question

5(4^(2x-1))=21

4^(2x-1)=21 / 5
=4.2

log[4^(2x-1)]=log4.2

Rule: log (a^x)=x loga

(2x-1) log4 = log4.2

2x-1 = log4.2 / log 4

x= [(log 4.2 / log 4) + 1] / 2

3. Originally Posted by calculus_jy
5(4^(2x-1))=21

4^(2x-1)=21 / 5
=4.2

log[4^(2x-1)]=log4.2

Rule: log (a^x)=x loga

(2x-1) log4 = log4.2

2x-1 = log4.2 / log 4

x= [(log 4.2 / log 4) + 1] / 2

Looks good to me. I could scream that...

-Dan

4. ## Thank you guys but....

See I knew I wasn't crazy I don't understand this at all....The answer is

ln21-ln5+ln4
2ln4

I have no idea. I got something with Log also so I know I'm not crazy.

Infact I got the exact thing you did.

5. Originally Posted by DEvans
See I knew I wasn't crazy I don't understand this at all....The answer is

ln21-ln5+ln4
2ln4

I have no idea. I got something with Log also so I know I'm not crazy.

Infact I got the exact thing you did.
First, you can use the log to any base you like, so you both have it right.

Also note that
log(4.2) = log(21/5) = log(21) - log(5)
(again, the log to any base) so this also works.

-Dan

6. ## should be rite

ln21-ln5+ln4
2ln4

here is how to get that

5[4^(2x+1)] = 21

4^(2x+1) = 21/5

ln[4^(2x+1)] = ln (21/5)

Rule:
log(a/b)= loga - logb

(2x+1)ln4 = (ln21 - ln 5)

(2x+1) = (ln21 - ln5) / ln4

2x = (ln21 - ln 5) / ln4 +1= (ln21 - ln5 + ln4) / ln4

x= ln21-ln5+ln4
2ln4

7. ## I've got it

I got confused thinking log was what I needed but if I start with a ln after dividing 21 by 5 then this is the result

ln(4^(2x-1)) = ln(21/5)

pull down (2x-1)

(2x-1)ln4 = ln(21/5)

divide by ln4

2x-1 = ln(21/5)/ln4

separate ln(21/5) into (ln21 - ln5)

2x-1 = (ln21 - ln5)/ln4

2x = [(ln21-ln5)+ln4]/ln4

divide by 2

x = (ln21-ln5+ln4)/2ln4

8. ## The other problem..

And it should be simple enough. I'm just not sure if I understand what's being asked.

Find the rule for the inverse of f if f(x) =

What do you guys think?

9. ## think this is rite

f(x) =

inverse function of f(x) is donated by f^-1 (x)

this operation is to isolate f^-1 (x) when x is replaced by f^-1 (x) and f(x) by x in the original function

thus

x = (2 . f^-1 (x) - 3) / ( f^-1 (x) - 5 )
x . f^-1 (x) - 5x = 2 . f^-1 (x) - 3
(x - 2) . f^-1 (x) = -3 + 5x
f^-1 (x) = (-3 + 5x) / (x - 2)

This should be rite!

10. Thanks I ended up getting it, what I needed to do was to subtract xy after I got the bottom cleared. Then after that factor it out. I was missing that vital step.

Thanks alot guys you've been a big help.

11. Originally Posted by DEvans
And it should be simple enough. I'm just not sure if I understand what's being asked.

Find the rule for the inverse of f if f(x) =

What do you guys think?
If you have a new question, please place it in a new thread.

-Dan