How should I solve these equations?

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February 21st 2011, 01:01 AM
mehdi

How should I solve these equations?

Hey all,
I am trying to solve these 2 equations but I can't.
$27^x + 12^x = 2 * 8^3$
and,
$3 * 16 ^x + 37 * 36^x = 26 * 81^x$
thanks in advance.
February 21st 2011, 01:30 AM
earboth

Quote:

Originally Posted by mehdi

Hey all,
I am trying to solve these 2 equations but I can't.
$27^x + 12^x = 2 * 8^3$
and,
$3 * 16 ^x + 37 * 36^x = 26 * 81^x$
thanks in advance.

I can show you how to do the 2nd one - but it is kind of tricky:

$3 * 16 ^x + 37 * 36^x = 26 * 81^x$

$37 \cdot 6^{2x}+3 \cdot 2^{4x} = 26 \cdot 3^{4x}$

$37 \cdot 2^{2x} \cdot 3^{2x}+3 \cdot 2^{4x} = 26 \cdot 3^{4x}$ divide by $2^{2x}$

$37 \cdot 3^{2x}+3 \cdot 2^{2x} = 26 \cdot 3^{4x} \cdot \dfrac1{2^{2x}}$ divide by $3^{2x}$

$37 +3 \cdot \left(\dfrac23\right)^{2x} = 26 \cdot 3^{2x} \cdot \dfrac1{2^{2x}} = 26 \cdot \left(\dfrac32\right)^{2x}$

Now substitute $y = \left(\dfrac32\right)^{2x}$ . The equation becomes:

$37+3 \cdot \dfrac1y=26 \cdot y$ multiply by y:

$26y^2-37y-3=0$

yields $y = \frac32~\vee~y=-\frac1{13}$

Since y is a power with a positive base the 2nd solution can't be correct.

Now re-substitute:

$\left(\dfrac32\right)^{2x} = \dfrac32$

That means $2x = 1~\implies~\boxed{x = \dfrac12}$
February 21st 2011, 01:31 AM
skoker

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February 21st 2011, 04:20 AM
Soroban

Hello, mehdi!

Another approach to the second problem . . .

Quote:

$3\cdot16^x + 37\cdot36^x \:=\: 26\cdot81^x$

We have: . $3(2^4)^x + 37(2^2\!\cdot\!3^3)^x - 26(3^4)^x \;=\;0$

. . . . . . . $3(2^{4x}) + 37(2^{2x})(3^{2x}) - 26(3^{4x}) \;=\;0$

Factor: . $\bigg[2^{2x} + 13\!\cdot\!3^{2x}\bigg]\,\bigg[3\!\cdot\!2^{2x} - 2\!\cdot\!3^{2x}\bigg] \;=\;0$

We have two equations to solve:

$2^{2x} + 13\!\cdot\!3^{2x} \:=\:0 \quad\Rightarrow\quad 2^{2x} \:=\:\text{-}13\!\cdot\!3^{2x}$

. . $\dfrac{2^{2x}}{3^{2x}} \:=\:\text{-}13 \quad\Rightarrow\quad \left(\dfrac{2}{3}\right)^{2x} \:=\:\text{-}13\;\hdots\;\text{ no real roots}$

$3\!\cdot\!2^{2x} - 2\!\cdot\!3^{2x} \:=\:0 \quad\Rightarrow\quad 3\!\cdot\!2^{2x} \:=\:2\!\cdot\!3^{2x}$

. . $\dfrac{2^{2x}}{3^{2x}} \:=\:\dfrac{2}{3} \quad\Rightarrow\quad \left(\dfrac{2}{3}\right)^{2x} =\:\left(\dfrac{2}{3}\right)^1 \quad\Rightarrow\quad 2x \:=\:1$

. . $\text{Therefore: }\:\boxed{x \:=\:\frac{1}{2}}$