Help with exponents/log

[PageOfHeart]

I need help with this equation:
(1/9)^(x+9)=3^(x) √527)^(10x^(2)-11)
I know it's a bit of a beast.
<3

 

[JeffM]

I need help with this equation:
(1/9)^(x+9)=3^(x) √527)^(10x^(2)-11)
I know it's a bit of a beast.
<3

You should know that putting anti-smiley faces into the middle of an equation is not terribly helpful.

Is this the equation to be solved

\(\displaystyle \left(\dfrac{1}{9}\right)^{(x + 9)} = 3^x\left(\sqrt{5 * 27}\right)^{(10x^2-11)}?\)

What we like to see is your work so we can know exactly where you need help. We also like to know where you are in your math education so that we can frame an answer that is appropriate for you. I doubt that suggesting Newton's method to you will be helpful, but I could be wrong. It also frequently helps if we know what you are currently studying.

The first thing I would try on this problem is converting both sides of the equation to log base 3. I am not going to try to work out an answer along that line because I do not know if I even have the problem right.

So please confirm what the problem to be solved is and give us some information that lets us help you intelligently.

 

[PageOfHeart]

I have fixed the smiley face issue. It's supposed to be 27 to the 5th root. I haven't done any work on it because I am not sure how to solve it in the first place.
<3

 

[JeffM]

I have fixed the smiley face issue. It's supposed to be 27 to the 5th root. I haven't done any work on it because I am not sure how to solve it in the first place.
<3

You can type out problems in plain text using PEMDAS, ^ to indicate exponentiation, * to indicate multiplication, / to indicate division, and fractional exponentiation to indicate roots.

You STILL have given us no information about your level of math education or what you are currently studying. I am GUESSING that you are taking second year high school algebra and that you are currently studying logarithms.

So I think the problem is

\(\displaystyle \left(\dfrac{1}{9}\right)^{(x + 9)} = 3^x * \left(\sqrt[5]{27}\right)^{(10x^2 - 11)}.\)

Is that correct?

A standard method of solving exponential equations is to use logarithms. That is why I asked if you were studying them. If I have now guessed correctly what the actual problem is, notice that 1/9, 3, and 27 are powers of three. That suggests using 3 as the base for the logarithms. Let's see if that idea works out. What do you get when, using base 3, you equate the logarithm of the left hand side of the equation to the logarithm of the right hand side of the equation

 

[DrPhil]

I need help with this equation:
(1/9)^(x+9)=3^(x) √527)^(10x^(2)-11)
I know it's a bit of a beast.
<3

One step toward the suggested approach of using \(\displaystyle \log_3\) is to replace (1/9) and 27 as powers of 3:

\(\displaystyle \displaystyle (3^{-2})^{x+9} = 3^x \left( ^5\sqrt{3^3} \right) ^{10x^2 - 11} \)

Now you need to apply what you know about rules for exponents. How far can you get?

 

[lookagain]

\(\displaystyle \displaystyle (3^{-2})^{x+9} = 3^x \left( ^5\sqrt{3^3} \right) ^{10x^2 - 11} \)

\(\displaystyle \sqrt[5]{3^3} \)


DrPhil,

if you would like the option, the fifth root of a quantity can be done in Latex by:

[ tex ] \sqrt[5]{quantity} [ \tex ] \(\displaystyle \ \ \ \ \ \) (without the spaces around "tex" and "\tex").