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?
$\displaystyle 3(2^{x}) = 18^{x-1}$
Take the natural log of both sides:
$\displaystyle ln(3(2^{x})) = ln(18^{x-1})$
Use the multiplication identity: ln(ab) = ln(a) + ln(b)
$\displaystyle ln(3)+ ln(2^{x}) = ln(18^{x-1})$
Use the exponent identity: $\displaystyle ln(a^{b}) = b\cdot ln(a)$
$\displaystyle ln(3)+ x\cdot ln(2) = (x-1)ln(18)$
Distribute the ln(18)
$\displaystyle ln(3)+ x\cdot ln(2) = x\cdot ln(18) - ln(18)$
Put all the terms with x in them on one side by subtracting x*ln(2) from both sides:
$\displaystyle ln(3) = x\cdot ln(18) - ln(18) - x\cdot ln(2)$
Put all the terms without x in them on the other side by adding ln(18) to both sides:
$\displaystyle ln(3) + ln(18) = x\cdot ln(18) - x\cdot ln(2)$
Factor out an x:
$\displaystyle ln(3) + ln(18) = x[ln(18) - ln(2)]$
Divide by the coefficient of x, which is [ln(18)-ln(2)]
$\displaystyle \frac{ln(3) + ln(18)}{ln(18) - ln(2)} = x$
Use additive identity of logarithms to see that $\displaystyle ln(3)+ln(18)=ln(3\cdot 18)=ln(54)$
$\displaystyle \frac{ln(54)}{ln(18) - ln(2)} = x$
Use subtraction identity to see that $\displaystyle ln(18)-ln(2)= ln(\frac{18}{2})=ln(9)$
$\displaystyle \frac{ln(54)}{ln(9)} = x$
Now you've solved for x, $\displaystyle x=\frac{ln(54)}{ln(9)}\approx 1.82$
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)
$\displaystyle 3 \cdot 2^x = 18^{x - 1}$
$\displaystyle \Rightarrow 3 \cdot 2^x = \left( 3^2 \cdot 2 \right)^{x - 1}$
$\displaystyle \Rightarrow 3 \cdot 2^x = 3^{2x - 2} \cdot 2^{x - 1}$
$\displaystyle \Rightarrow 3 \cdot 2^x = \frac 1{18} 3^{2x} \cdot 2^x$ .........2^x is never zero so we can divide by it
$\displaystyle \Rightarrow 3 = \frac {3^{2x}}{18}$
$\displaystyle \Rightarrow 3^{2x} = 3 \cdot 18 = 54$
here is where i said, "Oh, darn it! I need logs!"
$\displaystyle \Rightarrow 2x \ln 3 = \ln 54$
$\displaystyle \Rightarrow x \ln 9 = \ln 54$
$\displaystyle \Rightarrow x = \frac {\ln 54}{\ln 9} \approx 1.82$ as angel.white said
For this one, use the same strategy Angel White and Jhevon did, by taking the log of both sides.
$\displaystyle \ln \left[ 2(27)^x \right] = \ln 9^{x+1}$,
Then manipulate the equation using what you know about logs.
$\displaystyle \ln 2 + x \ln 27 = (x+1) \ln 9$.
Now no more x's are exponents, so you can just rearrange this equation using more elementary techniques to isolate the x.
yeah, crazy people see things sometimes. i'd also do some manipulations for the second question before applying logs as well
$\displaystyle 2 \cdot 27^x = 9^{x + 1}$
$\displaystyle \Rightarrow 2 \cdot 3^{3x} = 3^{2x + 2}$ .......3^(3x) is never zero, so we can divide by it
$\displaystyle \Rightarrow 2 = \frac {3^{2x + 2}}{3^{3x}}$
$\displaystyle \Rightarrow 2 = 3^{2 - x}$
now apply logs, the algebra of solving for x will be a lot simpler now
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