solve for x
8(5^2x) + 8(5^x) = 6
ALSO!
(3^5x)(9^x^2)= 27 solve for x
#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. $\displaystyle 2x^2 + 5x = 3$
Hi -
$\displaystyle 5^{2x} = 5^{x+x}=5^x \times 5^x=(5^x)^2$
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 $\displaystyle x$.
Re-write everything as a power of 3:(3^5x)(9^x^2)= 27 solve for x
$\displaystyle 3^{5x}.(3^2)^{x^2}=3^3$
$\displaystyle \implies 3^{5x}.3^{2x^2}=3^3$
... 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.
Grandad