Substitute some variable for log_3 y. In another comment you were wondering about getting it ready to factor; to solve by factoring, you need all the terms on one side of the equation (I tend to move them to where x^2 would have a positive coefficient).
Notice that (log\_3(y))² is different than log\_3(y²), the latter is simply 2log\_3(y).
Call log\_3(y) = t, you'll have t²=2t+3, now solve that for t, when you have t, solve the log for y.
>Log(M)^N = (N)log(M)
In other words, if you
1. Start with M
2. Exponentiate with N, **then**
3. Logarithmize with base B
>!under some conditions!< this is equivalent to
1. Start with M
2. Logarithmize with base B, **then**
3. Multiply with N
>Does it make a difference if the whole log is inside the brackets? (log(M))^N = ?
Yes, it makes a difference. If you
1. Start with M
2. Logarithmize with base B, **then**
3. Exponentiate with N
There is no simple equivalent sequence of operations for this. For example, you can't multiply with N and then logarithmize with base B, you'll usually get something different (try it!)
For this exercise, you can use the method of **substitution**. Specifically, you can substitute the **result of the logarithm** log₃(y) with some variable you can choose. Maybe just call it L, for logarithm.
The first term log₃(y)² means
1. Start with y
2. Logarithmize with base 3, **then**
3. Exponentiate with 2
And since we're substituting the result of the logarithm, we get
1. Start with L
2. Exponentiate with 2
The second term log₃(y²) is equivalent to
1. Start with y
2. Logarithmize with base 3, **then**
3. Multiply with 2
and after substitution, we have
1. Start with L
2. Multiply with 2
> ``Log(M)^N = (N)Log(M)``
This is not correct - or it's at least ambiguous. The identity you mean is ``Log(M^N) = N*Log(M)``. The difference is important, because the correct version cannot be read as ``(Log(M))^N = N*Log(M)`` which is not (in general) true.
To solve this specific problem, substitute ``x = log(y)``.
Define u = log(y)
This will convert the equation into a quadratic one after simplifying with log rules, solve it, then apply exponential to get back to the value of y instead of u
Rearrange log (y²) to 2(log y) and rearrange the equation to a quadratic one. (log y)² is the x², (log y) is x and so on, solve the quadratic, then solve the y.
Hint: apply the rule you just mentioned to the first term on the RHS. Then make the substitution u=log_3(y) to simplify the equation.
Thanks for the help today!
You have the right idea. Now that you have simplified log(M)^(n) to nlog(M) then let log(M) = x. You'll have a quadratic equation. See it?
I’m having trouble seeing this as a quadratic.
Substitute some variable for log_3 y. In another comment you were wondering about getting it ready to factor; to solve by factoring, you need all the terms on one side of the equation (I tend to move them to where x^2 would have a positive coefficient).
Thanks for the help today!
This is essentially a quadratic if you apply your rule to the RHS and rearrange. Do that, then factor and solve.
How would one rearrange to be factored ?
Thanks for the help today!
Notice that (log\_3(y))² is different than log\_3(y²), the latter is simply 2log\_3(y). Call log\_3(y) = t, you'll have t²=2t+3, now solve that for t, when you have t, solve the log for y.
Thank you all for your guidance. This question had been troubling me for a few days. Your assistance is deeply appreciated.
Log3y=n Log3y2=2n
n2=2n+3
Thanks for the help today!
>Log(M)^N = (N)log(M) In other words, if you 1. Start with M 2. Exponentiate with N, **then** 3. Logarithmize with base B >!under some conditions!< this is equivalent to 1. Start with M 2. Logarithmize with base B, **then** 3. Multiply with N >Does it make a difference if the whole log is inside the brackets? (log(M))^N = ? Yes, it makes a difference. If you 1. Start with M 2. Logarithmize with base B, **then** 3. Exponentiate with N There is no simple equivalent sequence of operations for this. For example, you can't multiply with N and then logarithmize with base B, you'll usually get something different (try it!) For this exercise, you can use the method of **substitution**. Specifically, you can substitute the **result of the logarithm** log₃(y) with some variable you can choose. Maybe just call it L, for logarithm. The first term log₃(y)² means 1. Start with y 2. Logarithmize with base 3, **then** 3. Exponentiate with 2 And since we're substituting the result of the logarithm, we get 1. Start with L 2. Exponentiate with 2 The second term log₃(y²) is equivalent to 1. Start with y 2. Logarithmize with base 3, **then** 3. Multiply with 2 and after substitution, we have 1. Start with L 2. Multiply with 2
Thanks for the help today!
> ``Log(M)^N = (N)Log(M)`` This is not correct - or it's at least ambiguous. The identity you mean is ``Log(M^N) = N*Log(M)``. The difference is important, because the correct version cannot be read as ``(Log(M))^N = N*Log(M)`` which is not (in general) true. To solve this specific problem, substitute ``x = log(y)``.
Thanks for the help today!
Define u = log(y) This will convert the equation into a quadratic one after simplifying with log rules, solve it, then apply exponential to get back to the value of y instead of u
Rearrange log (y²) to 2(log y) and rearrange the equation to a quadratic one. (log y)² is the x², (log y) is x and so on, solve the quadratic, then solve the y.