System of 3 logarithm equations: Find w: logx(yw) = 2 logy(zw) = 3 logz(xw) = 5

[TheAmazing]

Hi,
I've received this question, among others I don't completely understand, in my homework:

x, y, z, and w are real numbers that (?will be valid for?):
log_x(y^w) = 2
log_y(z^w) = 3
log_z(x^w) = 5
Find w.​

I've tried, but I just can't figure it out..
I do know however you can convert it to this system:

w*log_x(y) = 2
w*log_y(z) = 3
w*log_z(x) = 5

But that doesn't really help me...
If you figure it out, try not to straight up give me the answer but rather show me the way to get there.
Many thanks!

 

[Deleted member 4993]

Hi,
I've received this question, among others I don't completely understand, in my homework:

Find w:
log_x(y^w) = 2
log_y(z^w) = 3
log_z(x^w) = 5

I've tried, but I just can't figure it out..
I do know however you can convert it to this system:

w*log_x(y) = 2
w*log_y(z) = 3
w*log_z(x) = 5

But that doesn't really help me...
If you figure it out, try not to straight up give me the answer but rather show me the way to get there.
Many thanks!

log_x(y) *log_y(z) = log_x(z) = 1/[log_z(x)]

 

[ksdhart2]

Hi,
I've received this question, among others I don't completely understand, in my homework:

Find w:
log_x(y^w) = 2
log_y(z^w) = 3
log_z(x^w) = 5

I've tried, but I just can't figure it out..
I do know however you can convert it to this system:

w*log_x(y) = 2
w*log_y(z) = 3
w*log_z(x) = 5

But that doesn't really help me...
If you figure it out, try not to straight up give me the answer but rather show me the way to get there.
Many thanks!

First off, thank you for sharing your work with us without having to be asked. It's very refreshing to see someone actually read and obey the rules (maybe the two exclamation marks in the Read Before Posting thread aren't enough? ). That aside, I think a good next step would be to use the change of base formulahttp://www.mathwords.com/c/change_of_base_formula.htm to rewrite each of the equations. For example, the first one can be rewritten as:

\(\displaystyle w \cdot \dfrac{ln(y)}{ln(x)} = 2\)

If you solve each equation for w, then you have three expressions you know are all equal to w. See where that takes you.

 

[TheAmazing]

First off, thank you for sharing your work with us without having to be asked. It's very refreshing to see someone actually read and obey the rules (maybe the two exclamation marks in the Read Before Posting thread aren't enough? ). That aside, I think a good next step would be to use the change of base formula to rewrite each of the equations. For example, the first one can be rewritten as:

\(\displaystyle w \cdot \dfrac{ln(y)}{ln(x)} = 2\)

If you solve each equation for w, then you have three expressions you know are all equal to w. See where that takes you.

I went ahead and did that, and now I have this:
\(\displaystyle w = 2 - \dfrac{ln(y)}{ln(x)} \)

\(\displaystyle w = 3 - \dfrac{ln(z)}{ln(y)} \)

\(\displaystyle w = 5 - \dfrac{ln(x)}{ln(z)} \)

Or if you prefer, this: \(\displaystyle w = 2 - \dfrac{ln(y)}{ln(x)} = 3 - \dfrac{ln(z)}{ln(y)} = 5 - \dfrac{ln(x)}{ln(z)} \)
But other than that, I haven't really learned how to work with natural logarithms, let alone e..

However, I've tried using

log_x(y) *log_y(z) = log_x(z) = 1/[log_z(x)]

and w turned out as 30. I don't know if that's correct though, so if you want I can post my solving steps.

EDIT: I've discovered another solution for w, but I don't know if the first one is a mistake or not. The other solution I got is \(\displaystyle \sqrt[3]{30} \)

 

[stapel]

...However, I've tried using and w turned out as 30. I don't know if that's correct though, so if you want I can post my solving steps.

EDIT: I've discovered another solution for w, but I don't know if the first one is a mistake or not. The other solution I got is \(\displaystyle \sqrt[3]{30} \)

You can check the solution to any "solving" problem by plugging it back into the original exercise. Does either one of your solution values work?

 

[TheAmazing]

You can check the solution to any "solving" problem by plugging it back into the original exercise. Does either one of your solution values work?

That's the problem... I would've plugged it back in, but then I need to find valid x, y, and z that work with w, and that can take a while..