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log_{5} (x+7)=1-log_{5}(x+3) log_{5} means log base 5
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\[\log_5(x+7) = 1-\log_5(x+3) \iff \log_5(x+7) = \log_5(5)-\log_5(x+3)\]\[\log_5(x+7) = \log_5\left(\frac{5}{x+3}\right) \iff x+7 = \frac{5}{x+3}\] \[\iff (x+3)(x+7) = 5 \iff x^2+10x+16 = 0\] shouldnt be too bad from there.
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