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Log subscript 7(8x+20)=Log subscript 7(x+6) . Solve for x
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If\[\Large \log_{7}(8x+20)=\log_{7}(x+6)\]then\[\Large 8x+20=x+6\]
Use the property of logarithms that \(b^{\log_b{a}} = a\)
So now how do you solve ?
Do you know how to solve 8x + 20 = x + 6 ?
got x=-2
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That's your answer
That's correct!
log subscript4(4x) + log subscript 4(x)= Log subscript 4(64)
solve for x
\[\log{a}+\log{b} = \log{(a*b)}\]
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So you can rewrite as \(\log_4{(4x*x)} = \log_4(64)\) or \(\log_4{(4x^2)} = \log_4(64)\) From there you should be able to solve as in the previous problem.
x=4
Let's check: \[\log_4{4(4)} + \log_4{4} = \log_4{64}\]\[\log_4 16 + \log_4 4 = \log_4 64\]\[2+1=3\checkmark\]
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