equation 5(5)^(1/2)=25^(-5x+9)

[kitcait08]

5(5)^(1/2)=25^(-5x+9)

learned this a long time ago and all of my notes were stolen out of my car when they took my bag would appreciate any help....thanks!

i think it may have something to do with logs or making them have a common base but not sure!

-kitcait

 

[skeeter]

Re: equation

some hints for you ...

left side of the equation ...
\(\displaystyle 5 \cdot 5^{\frac{1}{2}} = 5^{\frac{3}{2}}\)

right side of the equation ...
\(\displaystyle 25^{-5x+9} = (5^2)^{-5x+9} = 5^{2(-5x+9)}\) edit dyslexic signs

general rule for exponential equations ...
if a[sup:302mmmxs]x[/sup:302mmmxs] = a[sup:302mmmxs]y[/sup:302mmmxs], the x = y.

 

[galactus]

Re: equation

You can use some property of logs tricks to make this fall into place.

\(\displaystyle 5\cdot\sqrt{5}=25^{-5x+9}\)

\(\displaystyle ln(5\cdot\sqrt{5})=(-5x+9)ln(25)\)

\(\displaystyle ln(\sqrt{125})=(-5x+9)ln(5^{2})\)

\(\displaystyle ln(\sqrt{5^{3}})=2(-5x+9)ln(5)\)

\(\displaystyle \frac{3}{2}ln(5)=2(-5x+9)ln(5)\)

\(\displaystyle \frac{3}{4}ln(5)=(-5x+9)ln(5)\)

Now, finish?. See what I done?. Now we have a common term, ln(5), and it's easier to deal with.

 

[Loren]

Re: equation

5(5)^(1/2)=25^(-5x+9)
\(\displaystyle 5\cdot 5^{\frac{1}{2}}=25^{-5x+9}\)

One approach is to rewrite with each side having the same base. Then the exponents are equal. That forms and equation which you can solve. In this case you might divide both sides by 5...
\(\displaystyle 5^{\frac{1}{2}}=\frac{25^{-5x+9}}{5^1}\)

\(\displaystyle 5^{\frac{1}{2}}=\frac{5^{2(-5x+9)}}{5^1}\)

\(\displaystyle 5^{\frac{1}{2}}=5^{2(-5x+9)-1}\)

\(\displaystyle \frac{1}{2}=2(-5x+9)-1\)

Now, solve for x, but before doing so, make sure you understand each of the steps leading up to this final line.

 

[Denis]

Re: equation

5(5)^(1/2)=25^(-5x+9)

Anutter way; square both sides:
25^(18-10x) = 125
18 - 10x = log(125) / log(25) ... carry on