nd-prabhakar
nd-prabhakar
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h^-1 (x) is y if h(y)=x So we need to write x=h(y)=ln(4-19y). Knowing x we want to solve for y (i.e., find y) ln(4-19y)=x means 4-19y=e^x, where e is the usual constant 2.718282 (approx) Subtract 4 from both sides -19y=e^x-4 Divide by -19 on…
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h^-1 (x) is y if h(y)=x So we need to write x=h(y)=ln(4-19y). Knowing x we want to solve for y (i.e., find y) ln(4-19y)=x means 4-19y=e^x, where e is the usual constant 2.718282 (approx) Subtract 4 from both sides -19y=e^x-4 Divide by -19 on…
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May be you are missing parentheses. The problem may be G(x)=6/(8-9x), in which G(x) is defined only if 8-9x is not 0, i.e., x is not equal to 8/9
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I am not sure what problem she finds in the statement. As far as I can see, it is correct: It is saying: (a^(km))^(1/kn) = a^(km/kn) using the rule that x^(M)^N=x^(M*N) =a^(m/n) canceling k which is okay since k is not 0 a^(km)^(1/kn) is…
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I am not sure what problem she finds in the statement. As far as I can see, it is correct: It is saying: (a^(km))^(1/kn) = a^(km/kn) using the rule that x^(M)^N=x^(M*N) =a^(m/n) canceling k which is okay since k is not 0 a^(km)^(1/kn) is…