Solve for y: Exponential Logarithm log8(9y+14)=6x6-11 | Step-by-Step Solution

[fr33pl4gu3]

log₈(9y+14)=6x⁶-11
9y+14=8^6x6-11
ln(9y+14)=ln8^6x6-11
ln9y+ln14=(6x⁶-11)ln8
ln9y=(6x⁶-11)ln8-ln14
y=((6x⁶-11)ln8-ln14)/ln9

is this equation correct??

 

[snipez90]

New Question:

log₈(9y + 14) = 6x⁶-11
9y+14 = 8^6x6-11

The next step is it by entering this step??

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

log₈(9y+14)=6x⁶-11
9y+14=8^6x6-11
ln(9y+14)=ln8^6x6-11
ln9y+ln14=(6x⁶-11)ln8
ln9y=(6x⁶-11)ln8-ln14
y=((6x⁶-11)ln8-ln14)/ln9

is this equation correct??

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.

 

[fr33pl4gu3]

log₈(9y+14)=6x⁶-11
9y+14=8^6x6-11
ln(9y+14)=ln8^6x6-11
ln126y=(6x⁶-11) ln8
y=((6x⁶-11) ln8)/ln126

Then is this correct??

 

[fr33pl4gu3]

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

 

[epkid08]

9y+14=8^6x6-11

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

 

[fr33pl4gu3]

9y + 14 = log8^6x6-log11
= log8^6x6 /log11
9y = log8^6x6 /log11 - 14
y = (1/9)log8^6x6 /log11 - (14 / 9)

Is this correct??

 

[HallsofIvy]

9y + 14 = log8^6x6-log11
= log8^6x6 /log11

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

9y = log8^6x6 /log11 - 14
y = (1/9)log8^6x6 /log11 - (14 / 9)

Is this correct??

 

[fr33pl4gu3]

9y + 14 = log8^6x6 -log11
= log8^6x6 /11
9y = (log8^6x6 /11) - 14
y = ((1/9)log8^6x6 /11) - (14 / 9)

Then, this would be correct, right??

 

[fr33pl4gu3]

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

 

[snipez90]

9y + 14 = log8^6x6 -log11
= log8^6x6 /11
9y = (log8^6x6 /11) - 14
y = ((1/9)log8^6x6 /11) - (14 / 9)

Then, this would be correct, right??

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!

 

[snipez90]

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

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

 

[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
8^{log₈(9y+14)} =8^6x6-11
9y+14=6x⁶-11
9y=6x⁶-25
y=(6x⁶-25)/9

 

[HallsofIvy]

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
8^{log₈(9y+14)} =8^6x6-11
9y+14=6x⁶-11

Yes, 8^{log_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.

9y=6x⁶-25
y=(6x⁶-25)/9