Need help!! test in 7 hours!!!

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December 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
December 17th 2008, 09:25 PM
o_O

#1:
$\begin{aligned} 6 & = 8 \cdot 5^{2x} + 8 \cdot 5^x \\ 0 & = 8 \left({\color{red}5^x}\right)^2 + 8 \cdot {\color{red}5^x} - 6 \\ 0 & = 4\left({\color{red}5^x}\right)^2 + 4\left({\color{red}5^x}\right) - 3\end{aligned}$

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

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

Factor to solve for $u$ .

__________

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

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

Since the bases are equal, then the exponents should be equal, i.e. $2x^2 + 5x = 3$
December 17th 2008, 09:28 PM
Grandad

Indices and logs

Hi -

Quote:

Originally Posted by capitalh

solve for x

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

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

So the equation can be re-writtenL

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

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

Quote:

(3^5x)(9^x^2)= 27 solve for x

Re-write everything as a power of 3:

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

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

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

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

Grandad
December 17th 2008, 10:03 PM
capitalh

thanks guys!