$\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)
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)} $