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Solve 2^x + ^3 = 7
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Something missing in your question
what is missing? I have to solve 2 to the power (x+3 = 7
oh \[2^{2x+3}=7\]
yes, exactlly
\[(x+3) = \log_{2}(7) \]
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sorry no, the two is only raised to the x+3 power
2^(x+3)=7 Use natural log ln(2^(x+3))=ln(7) xln(2)+3ln(2)=ln(7) xln(2)=-3ln(2)+ln(7) x=-(3ln(2))/(ln(2))+(ln(7))/(ln(2)) x=(-3ln(2)+ln(7))/(ln(2))_ x=-0.1926
\[x = \log_{2}(7)-3 \]
how do i solve Solve 11.32^(x – 1) = 15.7
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