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When and how do you use logarithms to solve exponential equations? give an example of an exponential equation that does not require logarithms to solve it and example of a problem that does require logarithms to solve it.

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Answer:

edge getting on my nerves

Explanation:

i need someone to explain this to me as well i do not get it at allll

User Leeks And Leaks
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\bf 5^(x+3)=\cfrac{1}{125}\implies 5^(x+3)=\cfrac{1}{5^3}\implies 5^(x+3)=5^(-3) \\\\\\ \textit{because the bases are the same, the exponents must also be the same} \\\\\\ x+3=-3\implies \boxed{x=-6}\\\\ -------------------------------\\\\ 3^x=4^(2x)\implies log(3^x)=log(4^(2x))\implies xlog(3)=(2x)log(4) \\\\\\ \cfrac{x}{2x}=\cfrac{log(4)}{log(3)}\implies \cfrac{1}{x}=\cfrac{log(4)}{log(3)}\implies \cfrac{log(3)}{log(4)}=x\implies \boxed{0.79248125\approx x}
User Surreal
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