Solve 9^(x+7) = 3^(3-x) without using logs

[jwpaine]

Have I done this right?

9^(x+7) = 3^(3-x)

these have different bases, so.... 3 to the (3-(x)) = 9? so I did 3^(3-1) = 9

so, 3^(3-x) = ((3)^2)^(x+7)

3^(3-x) = 3^(2x+7)

now each have the same base, so I solve for the x, exponent value:

3 - x = 2x + 7
-4 = 3x
x= (-4/3)

What have I done wrong?

 

[galactus]

\(\displaystyle \L\\9^{x+7}=(3^{2})^{x+7}=3^{2x+\overbrace{14}^{\text{here}}}\)

You have:

\(\displaystyle \L\\3^{2x+14}=3^{3-x}\)

Now, it's easy, ain't it?.

 

[jwpaine]

2x + 14 = 3 - x
3x = -11
x = (-11/3)

Thank you Galactus.



Now for this one...lets see if I get this right:


\(\displaystyle \L 16^{2x - 3} = 8^x\)

\(\displaystyle \L 16^{2x - 3} = (8^{4/3})^{2x - 3} = 8^{\frac{8x-12}{3}}\)

\(\displaystyle \L 8^{\frac{8x-12}{3}} = 8^x\)

Solve for X

\(\displaystyle \L 3(\frac{8x-12}{3}) = x(3)\)

\(\displaystyle \L 8x-12 = 3x\)

\(\displaystyle \L x = (12/5)\)

Is this good?

 

[galactus]

Yep, that's correct. But, if I may nit-pick. Perhaps this way would've been a wee bit easier.

\(\displaystyle \L\\16^{2x-3}=(2^{4})^{2x-3}=2^{8x-12}\)

Then,

\(\displaystyle \L\\2^{8x-12}=2^{3x}\)

\(\displaystyle \L\\8x-12=3x\)

\(\displaystyle \L\\x=\frac{12}{5}\)

Just a thought. I noticed you used fractional exponents.

You have the correct answer, nonetheless. Good work.

 

[jwpaine]

How would I do this one?

4^(2x-7) = 64

do I raise 4 to 4^3 = 64
or do I lower 64 to 64^(1/3) = 4

so they both have the same base.


using logs I could do

2x-7 = log(64)/log(4)
x = 5

because a^x = b: x = log(b)/log(a)

but how would I do it the way we were doing it before?

 

[skeeter]

take the easier route ... change 64 to 4³, then equate exponents and solve.

 

[galactus]

You could try it this way:

\(\displaystyle \L\\4^{2x-7}=64\)

\(\displaystyle \L\\4^{2x-7}=4^{3}\)

\(\displaystyle \L\\2x-7=3\)


Let's do something wacky:

\(\displaystyle \L\\4^{2x-7}=4^{2x}\cdot{4^{-7}}=\frac{16^{x}}{16384}\)

So, \(\displaystyle \L\\\frac{16^{x}}{16384}=64\)

\(\displaystyle \L\\16^{x}=1048576\)

\(\displaystyle \L\\16^{x}=16^{5}\)

\(\displaystyle \L\\x=5\)

 

[jwpaine]

Thanks Skeeter.

Thanks galactus.