Simplifying Logarithmic Ratios: How Do You Do It?

[casanova2528]

how the heck do you simplify this ?

(Ln (2x/y) / Ln (x/y)) = m/n

HELP ME!

 

[marcusl]

Start by cross-multiplying, then apply what you know about the log of an exponential expression.

 

[sutupidmath]

how the heck do you simplify this ?

(Ln (2x/y) / Ln (x/y)) = m/n

HELP ME!

I am just going to elaborate a lill bit what marcusl already suggested.

You probbably know that

log\frac{x}{y}=log(x)-log(y)

Also

log(ab)=log(a)+log(b)

just apply these properties, and yu'll be fine.

 

[casanova2528]

Ln (2x/y) = Ln 2x - Ln y

Ln (x/y) = Ln x - Ln y

[Ln (2x/y) / Ln (x/y)] = (Ln 2x - Ln y) / (Ln x - Ln y)

what do I do now?

 

[Gib Z]

Well that wasn't how I interpreted the original hint. After cross-multiplying as already said, to n \log_e \left( \frac{2x}{y} \right) = m \log_e \left( \frac{x}{y} \right), I would have applied the exponential function to both sides and simplified.

 

[sutupidmath]

Hell yeah.Gib Z is so right, my bad!

 

[casanova2528]

Well that wasn't how I interpreted the original hint. After cross-multiplying as already said, to n \log_e \left( \frac{2x}{y} \right) = m \log_e \left( \frac{x}{y} \right), I would have applied the exponential function to both sides and simplified.

that's not where I want to go.

basically, this natural log ratio reduces down to

1+ (ln 2)/Ln (X/Y)


how do you get here?

 

[uart]

casanova2528, you started with an equation (note the equals sign), so I assume you meant to write : 1+ (ln 2)/Ln (X/Y) = m/n.

To get this you should use the property of logs that ln(2x/y) = ln(2) + ln(x/y). You should find it pretty easy from there.

 

[casanova2528]

casanova2528, you started with an equation (note the equals sign), so I assume you meant to write : 1+ (ln 2)/Ln (X/Y) = m/n.

To get this you should use the property of logs that ln(2x/y) = ln(2) + ln(x/y). You should find it pretty easy from there.

thanks! Those darn properties!