K kpx001 Junior Member Joined Mar 6, 2006 Messages 119 Mar 16, 2006 #1 simplifying log7sqroot[x^4y^7] log7x^2(y) 2log7x+log7y thats what i got but its wrong. the log base needs to b 2. can anyone show step / process?
simplifying log7sqroot[x^4y^7] log7x^2(y) 2log7x+log7y thats what i got but its wrong. the log base needs to b 2. can anyone show step / process?
S soroban Elite Member Joined Jan 28, 2005 Messages 5,584 Mar 16, 2006 #2 Hello, kpx001! Simplify: \(\displaystyle \:\log_7\sqrt{x^4y^7}\) Click to expand... We have: \(\displaystyle \,\log_7\left(x^4y^7\right)^{\frac{1}{2}}\;=\;\frac{1}{2}\cdot\log\left(x^4y^7\right) \;= \;\frac{1}{2}\left[\log\left(x^4\right)\,+\,\log_7\left(y^7\right)\right] \;=\;\frac{1}{2}\left[4\cdot\log_7(x)\,+\,7\cdot\log_7(y)\right]\) Answer: \(\displaystyle \,2\cdot\log_7(x)\,+\,\frac{7}{2}\cdot\log_7(y)\) The log base needs to be 2. Click to expand... Now you tell us! . . . Use the Base-Change Formula. Answer: \(\displaystyle \L\,2\cdot\frac{\log_2(x)}{\log_2(7)}\,+\,\frac{7}{2}\cdot\frac{\log_2(y)}{\log_2(7)}\)
Hello, kpx001! Simplify: \(\displaystyle \:\log_7\sqrt{x^4y^7}\) Click to expand... We have: \(\displaystyle \,\log_7\left(x^4y^7\right)^{\frac{1}{2}}\;=\;\frac{1}{2}\cdot\log\left(x^4y^7\right) \;= \;\frac{1}{2}\left[\log\left(x^4\right)\,+\,\log_7\left(y^7\right)\right] \;=\;\frac{1}{2}\left[4\cdot\log_7(x)\,+\,7\cdot\log_7(y)\right]\) Answer: \(\displaystyle \,2\cdot\log_7(x)\,+\,\frac{7}{2}\cdot\log_7(y)\) The log base needs to be 2. Click to expand... Now you tell us! . . . Use the Base-Change Formula. Answer: \(\displaystyle \L\,2\cdot\frac{\log_2(x)}{\log_2(7)}\,+\,\frac{7}{2}\cdot\frac{\log_2(y)}{\log_2(7)}\)
