# Need help!! test in 7 hours!!!

• Dec 17th 2008, 09:02 PM
capitalh
Need help!! test in 7 hours!!!
solve for x

8(5^2x) + 8(5^x) = 6

ALSO!

(3^5x)(9^x^2)= 27 solve for x
• Dec 17th 2008, 09:25 PM
o_O
#1:
\displaystyle \begin{aligned} 6 & = 8 \cdot 5^{2x} + 8 \cdot 5^x \\ 0 & = 8 \left({\color{red}5^x}\right)^2 + 8 \cdot {\color{red}5^x} - 6 \\ 0 & = 4\left({\color{red}5^x}\right)^2 + 4\left({\color{red}5^x}\right) - 3\end{aligned}

This is basically a quadratic. To see it more easily, let $\displaystyle u = 5^x$.

Then our quadratic becomes: $\displaystyle 0 = 4u^2 + 4u - 3$

Factor to solve for $\displaystyle u$ .

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#2: Is this your question: $\displaystyle \left(3^{5x}\right)\left(9^{x^2}\right) = 27$

If so, try to get everything under the same base using your exponent laws:
\displaystyle \begin{aligned} \left(3^{5x}\right)\left( (3^2)^{x^2}\right) & = 27 \\ \left(3^{5x}\right)\left(3^{2x^2}\right) & = 3^3 \\ 3^{2x^2 + 5x} & = 3^3 \end{aligned}

Since the bases are equal, then the exponents should be equal, i.e. $\displaystyle 2x^2 + 5x = 3$
• Dec 17th 2008, 09:28 PM
Indices and logs
Hi -

Quote:

Originally Posted by capitalh
solve for x

8(5^2x) + 8(5^x) = 6

$\displaystyle 5^{2x} = 5^{x+x}=5^x \times 5^x=(5^x)^2$

So the equation can be re-writtenL

$\displaystyle 8(5^x)^2+8(5^x)-6=0$

Substitute $\displaystyle u=5^x$ to get a quadratic in $\displaystyle u$. Solve and then use $\displaystyle \text{ln }u=x \text{ln }5$ to find $\displaystyle x$.

Quote:

(3^5x)(9^x^2)= 27 solve for x
Re-write everything as a power of 3:

$\displaystyle 3^{5x}.(3^2)^{x^2}=3^3$

$\displaystyle \implies 3^{5x}.3^{2x^2}=3^3$

... can you do the next step? [Hint: $\displaystyle 3^a.3^b=3^{a+b}$]

You then get a quadratic in $\displaystyle x$ which you can solve.