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How do I solve this logarithm?
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\[\log _{7}(x+1) + \log _{7}(x-5) = 1\]
have you tried this ? basically, first step would be to use the property \(\large\log_ax+\log_ay=\log_axy\)
so that would give me\[\log _{7}(x+1)(x-5)\]or\[\log _{7}(x ^{2}-4x-5)\]right?
absolutely correct! thats your left side. now by the definition of logarithm, you can use : \(\large if, \log_ab=c, \implies b=a^c\)
so, what can you say about \(\log _{7}(x ^{2}-4x-5)=1\)
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I'm not sure how to do that... am I supposed to solve for x at any point?
what up what your face man
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