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Mathematics 16 Online
OpenStudy (anonymous):

Find all the real-number roots of the equation. Give an exact expression for the root and also a calculator approximation rounded to three decimal places. 10^2^x + 2(10^x) − 63 = 0

OpenStudy (anonymous):

To clarify: \[10^{2x}+2(10^{x)}-63=0\]

OpenStudy (anonymous):

the program that I am using didn't take that answer. Instead I guess it was looking for the answer to be \[\log _{10} (7)\]\[\approx.845\]

OpenStudy (anonymous):

I was being stupid. Sorry: let 10^x = y Therefore:\[10^{2x} + 2*10^x - 63 = y^2 +2y - 63 = 0\] Now use the quadratic formula and get that y = 7 or y = -9 Since we can't have log of a negative number: \[x = \log_{10}7 \]

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