How do you solve $3.5^(x+2) = 1.75^(x+3)$ ?

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Answer 1 · 2015-10-05T18:19:57

$x = (ln(7/16))/ln(2) approx -1.19265$

Take the natural logarithm (or any logarithm) of both sides and use properties of logarithms to help you solve for $x$ .

$ln(3.5^{x+2})=ln(1.75^{x+3})\rightarrow (x+2)ln(3.5)=(x+3)ln(1.75)$

$\rightarrow x(ln(1.75)-ln(3.5))=2ln(3.5)-3ln(1.75)$

$\rightarrow x=(ln(3.5^2)-ln(1.75^3))/(ln(1.75/3.5))=ln(16/7)/ln(1/2)$

$=ln(7/16)/ln(2) approx -1.19265$

How do you solve 3.5^(x+2) = 1.75^(x+3)3.5^(x+2) = 1.75^(x+3)?