exponetial equations

**imthatgirl** · November 6th 2007, 12:25 PM

Solve in exact form

1) 3(2)^x = 18^(x-1)

2) 2(27)^x = 9^(x+1)

Can anyone help with any of these questions?

**angel.white** · November 6th 2007, 01:04 PM

Originally Posted by imthatgirl

Solve in exact form

1) 3(2)^x = 18^(x-1)

2) 2(27)^x = 9^(x+1)

Can anyone help with any of these questions?

$3(2^{x}) = 18^{x-1}$

Take the natural log of both sides:
$ln(3(2^{x})) = ln(18^{x-1})$

Use the multiplication identity: ln(ab) = ln(a) + ln(b)
$ln(3)+ ln(2^{x}) = ln(18^{x-1})$

Use the exponent identity: $ln(a^{b}) = b\cdot ln(a)$
$ln(3)+ x\cdot ln(2) = (x-1)ln(18)$

Distribute the ln(18)
$ln(3)+ x\cdot ln(2) = x\cdot ln(18) - ln(18)$

Put all the terms with x in them on one side by subtracting x*ln(2) from both sides:
$ln(3) = x\cdot ln(18) - ln(18) - x\cdot ln(2)$

Put all the terms without x in them on the other side by adding ln(18) to both sides:
$ln(3) + ln(18) = x\cdot ln(18) - x\cdot ln(2)$

Factor out an x:
$ln(3) + ln(18) = x[ln(18) - ln(2)]$

Divide by the coefficient of x, which is [ln(18)-ln(2)]
$\frac{ln(3) + ln(18)}{ln(18) - ln(2)} = x$

Use additive identity of logarithms to see that $ln(3)+ln(18)=ln(3\cdot 18)=ln(54)$
$\frac{ln(54)}{ln(18) - ln(2)} = x$

Use subtraction identity to see that $ln(18)-ln(2)= ln(\frac{18}{2})=ln(9)$
$\frac{ln(54)}{ln(9)} = x$

Now you've solved for x, $x=\frac{ln(54)}{ln(9)}\approx 1.82$

**angel.white** · November 6th 2007, 01:12 PM

The second equation will be $x \approx -2.63$ can you figure out what the exact form is?

**Jhevon** · November 6th 2007, 01:18 PM

an alternate approach. (i originally thought it could be done without logs, so i tried simplifying to equate like powers, but i had to end up using them at the end)

$3 \cdot 2^x = 18^{x - 1}$

$\Rightarrow 3 \cdot 2^x = \left( 3^2 \cdot 2 \right)^{x - 1}$

$\Rightarrow 3 \cdot 2^x = 3^{2x - 2} \cdot 2^{x - 1}$

$\Rightarrow 3 \cdot 2^x = \frac 1{18} 3^{2x} \cdot 2^x$ .........2^x is never zero so we can divide by it

$\Rightarrow 3 = \frac {3^{2x}}{18}$

$\Rightarrow 3^{2x} = 3 \cdot 18 = 54$

here is where i said, "Oh, darn it! I need logs!"

$\Rightarrow 2x \ln 3 = \ln 54$

$\Rightarrow x \ln 9 = \ln 54$

$\Rightarrow x = \frac {\ln 54}{\ln 9} \approx 1.82$ as angel.white said

**angel.white** · November 6th 2007, 01:27 PM

Originally Posted by Jhevon

an alternate approach. (i originally thought it could be done without logs, so i tried simplifying to equate like powers, but i had to end up using them at the end)

$3 \cdot 2^x = 18^{x - 1}$

$\Rightarrow 3 \cdot 2^x = \left( 3^2 \cdot 2 \right)^{x - 1}$

$\Rightarrow 3 \cdot 2^x = 3^{2x - 2} \cdot 2^{x - 1}$

$\Rightarrow 3 \cdot 2^x = \frac 1{18} 3^{2x} \cdot 2^x$ .........2^x is never zero so we can divide by it

$\Rightarrow 3 = \frac {3^{2x}}{18}$

$\Rightarrow 3^{2x} = 3 \cdot 18 = 54$

here is where i said, "Oh, darn it! I need logs!"

$\Rightarrow 2x \ln 3 = \ln 54$

$\Rightarrow x \ln 9 = \ln 54$

$\Rightarrow x = \frac {\ln 54}{\ln 9} \approx 1.82$ as angel.white said

I'm impressed that you thought of that, even if I seen that, I think I would have gotten too lost after just a few steps, and given up lol, gj.

**Soltras** · November 6th 2007, 01:38 PM

Originally Posted by imthatgirl

2) 2(27)^x = 9^(x+1)

For this one, use the same strategy Angel White and Jhevon did, by taking the log of both sides.

$\ln \left[ 2(27)^x \right] = \ln 9^{x+1}$ ,

Then manipulate the equation using what you know about logs.

$\ln 2 + x \ln 27 = (x+1) \ln 9$ .

Now no more x's are exponents, so you can just rearrange this equation using more elementary techniques to isolate the x.

**Jhevon** · November 6th 2007, 01:39 PM

Originally Posted by angel.white

I'm impressed that you thought of that, even if I seen that, I think I would have gotten too lost after just a few steps, and given up lol, gj.

yeah, crazy people see things sometimes. i'd also do some manipulations for the second question before applying logs as well

$2 \cdot 27^x = 9^{x + 1}$

$\Rightarrow 2 \cdot 3^{3x} = 3^{2x + 2}$ .......3^(3x) is never zero, so we can divide by it

$\Rightarrow 2 = \frac {3^{2x + 2}}{3^{3x}}$

$\Rightarrow 2 = 3^{2 - x}$

now apply logs, the algebra of solving for x will be a lot simpler now

**imthatgirl** · November 6th 2007, 06:27 PM

Thanks so much!!! and yeah i get it now, but i did the second one and i got a different answer, i actually got

log(2/9) / log(1/3) = 1.369
and when i substitute it back in, it worked... but i'll compare answers with someone tomorrow...
but thanks so much for helping

**Jhevon** · November 6th 2007, 06:53 PM

Originally Posted by imthatgirl

Thanks so much!!! and yeah i get it now, but i did the second one and i got a different answer, i actually got

log(2/9) / log(1/3) = 1.369
and when i substitute it back in, it worked... but i'll compare answers with someone tomorrow...
but thanks so much for helping

that is the correct answer. what method did you use to arrive at it?

**angel.white** · November 6th 2007, 09:40 PM

Originally Posted by imthatgirl

Thanks so much!!! and yeah i get it now, but i did the second one and i got a different answer, i actually got

log(2/9) / log(1/3) = 1.369
and when i substitute it back in, it worked... but i'll compare answers with someone tomorrow...
but thanks so much for helping

That is correct, my apologies, I transcribed the right side to my paper as $9^{x-1}$ so I solved for a different equation

**imthatgirl** · November 10th 2007, 07:44 PM

I used angel.white's method
but thanks for the help,

and no worries angel.white, u explained really well for the first one =D

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