1. ## Logs

$\displaystyle y = 6* 5^x$and $\displaystyle y = 9^x$ intersect at the point Q. Show that the x-coordinate of Q can be
written as x = (1 + log3 2) / (2 − log3 5)

2. Originally Posted by BabyMilo
y = 6 × 5x and y = 9x intersect at the point Q. Show that the x-coordinate of Q can be
written as x = (1 + log3 2) / (2 − log3 5)
The question as written only has the solution (0,0)

This is because we have $\displaystyle 30x=9x$.

Are you sure you typed it correctly?

3. Originally Posted by e^(i*pi)
The question as written only has the solution (0,0)

This is because we have $\displaystyle 30x=9x$.

Are you sure you typed it correctly?
sorry, corrected.

4. and
logy = log6 + xlog5 .....(1)
logy = xlog9 ...........(2)
So
xlog9 =log6 +xlog5
Solved for x and change the base to 3.

5. Hello, BabyMilo!

You really must learn to write exponential expressions . . .

$\displaystyle y \,=\, 6\cdot5^x$ and $\displaystyle y \,=\, 9^x$ intersect at point $\displaystyle Q.$

Show that the $\displaystyle x$-coordinate of $\displaystyle Q$ is: .$\displaystyle x \;=\;\frac{1+\log_32}{2 - \log_35}$

We have: .$\displaystyle 9^x \:=\:6\cdot5^x$

Take logs (base 3): .$\displaystyle \log_3\left(9^x\right) \;=\;\log_3\left(6\cdot5^x\right)$

. . . . . . . . . . . . . . $\displaystyle \log_3(3^2)^x \;=\;\log_3(6) + \log_3(5^x)$

. . . . . . . . . . . . . . $\displaystyle \log_3(3^{2x}) \;=\;\log_3(3\cdot2) + \log_3(5^x)$

. . . . . . . . . . . . . .$\displaystyle 2x\underbrace{\log_3(3)}_{\text{This is 1}} \;=\;\underbrace{\log_3(3)}_{\text{This is 1}} + \log_3(2) + x\log_3(5)$

. . . . . . . . . . . . . . . . . .$\displaystyle 2x \;=\;1 + \log_3(2) + x\log_3(5)$

. . . . . . . . . . . $\displaystyle 2x - x\log_3(5) \;=\;1 + \log_3(2)$

Factor: . . . . . $\displaystyle x\bigg[2 - \log_3(5)\bigg] \;=\;1 + \log_3(2)$

Therefore: . . . . . . . . . . .$\displaystyle x \;=\;\frac{1 + \log_3(2)}{2 - \log_3(5)}$