Solve for $x$: $14^{4x} = 3^{-x-3}$ Write the exact answer using base-10 logarithms

Question

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Solve for x. $$14^{4x} = 3^{-x-3}$$ Write the exact answer using base-10 logarithms

Answer 1

$$14^{4x} = 3^{-x-3}$$

$$ \Leftrightarrow \log(14^{4x}) = \log(3^{-x-3})$$

$$ \Leftrightarrow 4x \log(14) = (-x-3) \log(3) $$

$$ \Leftrightarrow 4x\log(14) = -x\log(3)-3\log(3) $$

$$ \Leftrightarrow 4x\log(14)+x\log(3) = -3\log(3) $$

$$ \Leftrightarrow x(4\log(14)+\log(3)) = -3\log(3) $$

$$ \Leftrightarrow x = \frac{-3\log(3)}{4\log(14)+\log(3)} $$

Which can be further simplified if desired.

Answer 2

$$14^{4x} = 3^{-x-3} \implies (14^{4})^x = 3^{-x}3^{-3} \implies (14^{4}\cdot 3)^x = 3^{-3}$$

$$\implies x =\frac{\log_{10}(3^{-3})}{\log_{10}\left(3\cdot 14^4\right)}$$

Answer 3

Hint

$$a^x=b^x\implies 10^{x\log(a)}=10^{x\log(b)}.$$

Answer 4

Hint:

Using base-10 logarithms the equation becomes: $$ 4x \log_{10} 14=(-x-3)\log_{10} 3 \quad \iff \quad x(4\log_{10}14+\log_{10}3)=-3\log_{10}3 $$

that is a simple first degree equation in $x$.