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

This is correct, but you should show your steps and explain your reasoning more clearly. Also, please use proper notation with parentheses and clear labeling of variables.
  • #1
fr33pl4gu3
82
0
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??
 
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  • #2
fr33pl4gu3 said:
New Question:

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

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?

fr33pl4gu3 said:
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??

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
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
When multiply the log sign, can i just multiply one side??
 
  • #5
fr33pl4gu3 said:
9y+14=86x6-11

You seem to be over complicating things. From here just do some simple algebra to solve for y.
 
  • #6
9y + 14 = log86x6-log11
= log86x6 /log11
9y = log86x6 /log11 - 14
y = (1/9)log86x6 /log11 - (14 / 9)

Is this correct??
 
  • #7
fr33pl4gu3 said:
9y + 14 = log86x6-log11
= log86x6 /log11
No. lod a- log b is not log a/log b. It is log a/b

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

Is this correct??
 
  • #8
9y + 14 = log86x6 -log11
= log86x6 /11
9y = (log86x6 /11) - 14
y = ((1/9)log86x6 /11) - (14 / 9)

Then, this would be correct, right??
 
  • #9
Usually, how does an equation have the condition of b(logbu) =u??
 
  • #10
fr33pl4gu3 said:
9y + 14 = log86x6 -log11
= log86x6 /11
9y = (log86x6 /11) - 14
y = ((1/9)log86x6 /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!
 
  • #11
fr33pl4gu3 said:
Usually, how does an equation have the condition of b(logbu) =u??

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
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
fr33pl4gu3 said:
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
Yes, 8log_8(9y+ 14)= 9y+ 14 but what happened to the base 8 on the right side?

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

9y=6x6-25
y=(6x6-25)/9
 
Last edited by a moderator:

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

What is an exponential logarithm?

An exponential logarithm is a mathematical function that is the inverse of an exponential function. It is used to solve equations with exponential variables.

How do I solve for y in an exponential logarithm equation?

To solve for y in an exponential logarithm equation, you need to use the properties of logarithms to isolate the variable. In this case, you would first use the property log_a(b^c) = c * log_a(b) to move the 6 outside of the logarithm, and then use the property log_a(b * c) = log_a(b) + log_a(c) to separate the terms inside the parentheses.

What is the base of a logarithm?

The base of a logarithm is the number that is raised to a power to get the input of the logarithm. In this equation, the base is 8 since the logarithm is log8.

How do I know if my solution for y is correct?

To check if your solution for y is correct, you can substitute it back into the original equation and see if it satisfies the equation. In this case, you would plug in the value of y that you solved for and see if the equation is true.

What is the step-by-step solution for this equation?

The step-by-step solution for this equation is as follows:
1. Use the property log_a(b^c) = c * log_a(b) to move the 6 outside of the logarithm: log8(9y+14) = 6 * log8(6-11)
2. Use the property log_a(b * c) = log_a(b) + log_a(c) to separate the terms inside the parentheses: log8(9y+14) = 6 * (log8(6) - log8(11))
3. Simplify the logarithms using the fact that log_a(a) = 1: log8(9y+14) = 6 * (log8(6) - 1)
4. Use the inverse property of logarithms to rewrite the equation as an exponential equation: 9y+14 = 8^(6 * (log8(6) - 1))
5. Simplify the exponent using the property a^(b * c) = (a^b)^c: 9y+14 = (8^6)^(log8(6) - 1)
6. Simplify the exponent further using the fact that 8^6 = 262144: 9y+14 = 262144^(log8(6) - 1)
7. Use the fact that 262144 = 8^5 to rewrite the equation as: 9y+14 = (8^5)^(log8(6) - 1)
8. Simplify the exponent again using the property a^(b * c) = (a^b)^c: 9y+14 = 8^(5 * (log8(6) - 1))
9. Use the inverse property of logarithms to rewrite the equation as: 9y+14 = 8^(5 * log8(6)) * 8^(-5)
10. Simplify the logarithm using the property log_a(a^b) = b: 9y+14 = 6 * 8^(-5)
11. Use the property a^(-b) = 1/(a^b) to simplify further: 9y+14 = 6/(8^5)
12. Simplify the right side of the equation: 9y+14 = 6/32768
13. Subtract 14 from both sides: 9y = -32758/32768
14. Simplify the fraction: 9y = -16379/16384
15. Divide both sides by 9: y = -16379/16384 * 1/9
16. Simplify the fraction: y = -16379/147456
17. This is the final solution for y.

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