# Homework Help: Exponential logarithm

1. Aug 23, 2008

### fr33pl4gu3

log8(9y+14)=6x6-11
9y+14=86x6-11
ln(9y+14)=ln86x6-11
ln9y+ln14=(6x6-11)ln8
ln9y=(6x6-11)ln8-ln14
y=((6x6-11)ln8-ln14)/ln9

is this equation correct??

2. Aug 23, 2008

### snipez90

You have y on the LHS, just solve for it. Why would you continue to take the logarithm?

That is not correct. ln(a+b) =/= ln(a) + ln(b). I think you should take another look at what a logarithm means and not a rule sheet.

3. Aug 23, 2008

### fr33pl4gu3

log8(9y+14)=6x6-11
9y+14=86x6-11
ln(9y+14)=ln86x6-11
ln126y=(6x6-11) ln8
y=((6x6-11) ln8)/ln126

Then is this correct??

4. Aug 23, 2008

### fr33pl4gu3

When multiply the log sign, can i just multiply one side??

5. Aug 23, 2008

### epkid08

You seem to be over complicating things. From here just do some simple algebra to solve for y.

6. Aug 23, 2008

### fr33pl4gu3

9y + 14 = log86x6-log11
= log86x6 /log11
9y = log86x6 /log11 - 14
y = (1/9)log86x6 /log11 - (14 / 9)

Is this correct??

7. Aug 23, 2008

### HallsofIvy

No. lod a- log b is not log a/log b. It is log a/b

8. Aug 23, 2008

### fr33pl4gu3

9y + 14 = log86x6 -log11
= log86x6 /11
9y = (log86x6 /11) - 14
y = ((1/9)log86x6 /11) - (14 / 9)

Then, this would be correct, right??

9. Aug 23, 2008

### fr33pl4gu3

Usually, how does an equation have the condition of b(logbu) =u??

10. Aug 23, 2008

### snipez90

Um, HallsofIvy explained why you CANNOT have that. Also please use parentheses and clear up the notation. You have no idea how frustrating it is to read logarithms without parentheses around the arguments, i.e. use the form ln(the expression inside of here).

Start with the line that epkid08 quoted you on and simply solve for y. Don't take anymore logarithms!

11. Aug 23, 2008

### snipez90

Ask yourself what logbu stands for. If your answer included the word exponent, then it's likely you'll immediately see that it holds by definition!

12. Aug 24, 2008

### fr33pl4gu3

The previous solution start off incorrectly, so i start off a new one, but there were all wrong, such as below:

log8(9y+14)=6x6-11
8log8(9y+14) =86x6-11
9y+14=6x6-11
9y=6x6-25
y=(6x6-25)/9

13. Aug 24, 2008

### HallsofIvy

Yes, 8log_8(9y+ 14)= 9y+ 14 but what happened to the base 8 on the right side?

$$9y+ 14= 8^{6x^6- 11}$$
That's easy to solve for y.

Last edited by a moderator: Aug 24, 2008