Author Topic: Logarithmic can exponential functions (Read 5068 times) Tweet Share

AppleXY · « **Reply #15 on:** April 12, 2008, 06:34:39 pm »

Hey purple, I see you're making silly mistakes. Check every step and remember the -s

purple_rose · « **Reply #16 on:** April 12, 2008, 06:36:42 pm »

yea i think i get most of it, thankzzzzzzzzzzzzzz guys

starbuckscoffee · « **Reply #17 on:** April 13, 2008, 05:39:25 pm »

1. The graph of the function with rule f(x)= a loge (-x)+b passes through the points (-2,6) and (-4,8). Find the value of b (i can get the a, it's 2/loge (2))

Thanks

Mao · « **Reply #18 on:** April 13, 2008, 06:50:22 pm »

you've already got a, just sub it back into the function:

given that $a=\frac{2}{ln(2)}$

$\implies f(x)=\frac{2}{ln(2)}\cdot ln(-x)+b$

$\implies f(-2)=\frac{2}{ln(2)}\cdot ln(2) +b = 6$

$\implies 2 + b = 6$

$\implies b=4$

starbuckscoffee · « **Reply #19 on:** April 13, 2008, 06:55:37 pm »

thanks.....oic what i did, i thought cause i already used (-2,6) to get 2/loge (2), i couldn't sub it in again

starbuckscoffee · « **Reply #20 on:** April 14, 2008, 07:02:43 pm »

how do you solve 2^1/2(x+2) = 3^(x-1) and 2^(x+1) * 3^(x-1)=100?
Thanks

Collin Li · « **Reply #21 on:** April 14, 2008, 07:06:00 pm »

$2^{x+1} \times 3^{x-1} = 100$

$\implies 2\left(2^x\right) \times \frac{3^x}{3} = 100$

$\implies \frac{2}{3}\left(2\times3\right)^x = 100$

$\implies 6^x = 150$

$\implies x = \log_6150$

To get that in a more convenient base, use a change of base formula, or the following manipulation (same as using the formula, but doesn't require you to remember the formula):

From $6^x = 150$ : $\implies x\log_e6 = \log_e150$

$x = \frac{\log_e150}{\log_e6} \approx 2.796$

(Of course, if you wanted it to decimal places, you'd just use your calculator's graphical capabilities or the solve() function to work it out, rather than doing the algebra).

Oops, I missed this one: $2^{\frac{1}{2}(x+2)} = 3^{x-1}$

$\implies \left(\sqrt{2}\right)^{x+2} = 3^{x-1}$

$\implies \left(\sqrt{2}\right)^2 \times \left(\sqrt{2}\right)^x = \frac{3^x}{3}$

$\implies 6 = \left(\frac{3}{\sqrt{2}}\right)^x$

$\implies \log_e6 = x\log_e\left(\frac{3}{\sqrt{2}}\right)$

$\implies x = \frac{\log_e6}{\log_e3 - 0.5\log_e2}$

starbuckscoffee · « **Reply #22 on:** April 14, 2008, 07:15:18 pm »

starbuckscoffee · « **Reply #23 on:** April 14, 2008, 07:19:03 pm »

Quote from: coblin on April 14, 2008, 07:06:00 pm

$\implies 6 = \left(\frac{3}{\sqrt{2}}\right)^x$

How did you get this?

Don't worry i c now

thanks coblin, always reliable!

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Author Topic: Logarithmic can exponential functions (Read 5068 times) Tweet Share

AppleXY

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