# Intersection of 2 log functions

• January 6th 2009, 02:11 PM
mike_302
Intersection of 2 log functions
So the log functions given are y=3(5^(2x)) and y=6(4^(3x)) .

We are asked to find their point of intersection. To do this, I set each one to equal the other, 3(5^(2x))=6(4^(3x)) , and solve to find the x-value that the two meet at, then I'll sub that in later to find the y-value....

But how do I solve this: 3(5^(2x))=6(4^(3x)) ....... I can't seem to figure that one out. We are doing logarithmic equations practice... We know the product, quotient and power laws of logs.

Thanks!
• January 6th 2009, 02:16 PM
craig
Quote:

Originally Posted by mike_302
So the log functions given are y=3(5^(2x)) and y=6(4^(3x)) .

We are asked to find their point of intersection. To do this, I set each one to equal the other, 3(5^(2x))=6(4^(3x)) , and solve to find the x-value that the two meet at, then I'll sub that in later to find the y-value....

But how do I solve this: 3(5^(2x))=6(4^(3x)) ....... I can't seem to figure that one out. We are doing logarithmic equations practice... We know the product, quotient and power laws of logs.

Thanks!

Sorry if I'm misunderstanding this, but what exactly are the logs?

Is it $\log_{3} 5^{2x}$ ?
• January 6th 2009, 02:17 PM
Chop Suey
$3(5^{2x})=6(4^{3x})$

If we divide by $3$ and $4^{3x}$, we get:

$\frac{5^{2x}}{4^{3x}} = 2$

$\implies \left(\frac{5^2}{4^3}\right)^x = 2$

Can you do it now using what you know with laws of logs?
• January 6th 2009, 02:31 PM
mike_302
sorry: Bad wording on my part... Chop Suey's got it though.. I'm just trying to work out how you went from your first step to the last.
• January 6th 2009, 02:34 PM
mike_302
Yep, I've figured it out... works out to -0.737 (approx) .

Thanks!
• January 6th 2009, 02:35 PM
Chop Suey
Quote:

Originally Posted by mike_302
sorry: Bad wording on my part... Chop Suey's got it though.. I'm just trying to work out how you went from your first step to the last.

One of the properties of exponents are:
$(ab)^m = a^m b^m$

In this case, we have $\frac{a^{mn}}{b^{on}} = \left(\frac{a^m}{b^o}\right)^n$
EDIT:
Quote:

Originally Posted by mike_302
Yep, I've figured it out... works out to -0.737 (approx) .

Thanks!

Good job. (Yes)