For this email I will use a convention where if a^x = b then x is written as log(a) b. i.e. the base of the log is written in parantheses. So if for example we have (ab)^c = d then c is given by log(ab) d.

Now I think that log(xy) z can be rewritten as following

I refer to <a href="/exponential-logarithmic-functions/3-logarithm-laws.php">3. Logarithm Laws</a>
For this email I will use a convention where if a^x = b then x is written as log(a) b. i.e. the base of the log is written in parantheses. So if for example we have (ab)^c = d then c is given by log(ab) d.
Now I think that log(xy) z can be rewritten as following
log(xy) z = 1/[(1/log(x) z) + (1/log(y) z)]
Have you heard of such an identity?

Relevant page
<a href="/exponential-logarithmic-functions/3-logarithm-laws.php">3. Logarithm Laws</a>
What I've done so far
I've been thinking about this for some time now.

It is quite hard to read your question. You are encouraged to use the math entry system.

Actually, I have not seen this before, but it is true. I am using change of base formula (which is on this page: 5. Natural Logarithms (base e)

I am changing to base 10, so I just write "log" (but I could change it to any base).

`\text{LHS}`

`= \log_{xy} z`

`= \frac {log z}{\log xy}`

`= \frac {log z}{\log x + \log y}`

`\text{RHS}`

`= \frac{1}{(\log x / \log z)} + \frac{\log y}{\log z}`

`= \frac{1}{(\log x + \log y)/\log z}`

`= \frac{\log z}{\log x + \log y}`

Phew!

I'm not sure what you would use it for, though!

Regards

X

Hi Michael
It is quite hard to read your question. You are encouraged to use the math entry system.
Actually, I have not seen this before, but it is true. I am using change of base formula (which is on this page: <a href="/exponential-logarithmic-functions/5-logs-base-e-ln.php">5. Natural Logarithms <span class="noWrap">(base e)</span></a>
I am changing to base 10, so I just write "log" (but I could change it to any base).
`\text{LHS}`
`= \log_{xy} z`
`= \frac {log z}{\log xy}`
`= \frac {log z}{\log x + \log y}`
`\text{RHS}`
`= \frac{1}{(\log x / \log z)} + \frac{\log y}{\log z}`
`= \frac{1}{(\log x + \log y)/\log z}`
`= \frac{\log z}{\log x + \log y}`
Phew!
I'm not sure what you would use it for, though!
Regards