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How do you solve $3^{- 3 p} = 3^{3 p}$ $3^(-3p)=3^(3p)$ ?

2 Answers

Konstantinos Michailidis

Feb 3, 2017

We have

$3^{- 3 p} = 3^{3 p}$ $3^(-3p)=3^(3p)$

$\frac{1}{3^{3 p}} = 3^{3 p}$ $1/(3^(3p))=3^(3p)$

$1 = 3^{3 p + 3 p}$ $1=3^(3p+3p)$

$1 = 3^{6 p}$ $1=3^(6p)$

$ln 1 = 6 \cdot p \cdot ln 3$ $ln1=6*p*ln3$ (take logarithms on both sides)

$0 = p \cdot (6 \cdot ln 3)$ $0=p*(6*ln3)$

$p = 0$ $p=0$

Hence $p = 0$ $p=0$

Konstantinos Michailidis

Feb 3, 2017

We have

$3^{- 3 p} = 3^{3 p}$ $3^(-3p)=3^(3p)$

$\frac{1}{3^{3 p}} = 3^{3 p}$ $1/(3^(3p))=3^(3p)$

$1 = 3^{3 p + 3 p}$ $1=3^(3p+3p)$

$1 = 3^{6 p}$ $1=3^(6p)$

$ln 1 = 6 \cdot p \cdot ln 3$ $ln1=6*p*ln3$ (take logarithms on both sides)

$0 = p \cdot (6 \cdot ln 3)$ $0=p*(6*ln3)$

$p = 0$ $p=0$

Hence $p = 0$ $p=0$

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