Z zee New member Joined Jan 15, 2006 Messages 8 Jan 21, 2006 #1 I don not know how to solve for y when there is a log only on one side of the equation. Please explain how to do this problem. Express y in terms of x. log y = 1.2x - 1 Thank you.
I don not know how to solve for y when there is a log only on one side of the equation. Please explain how to do this problem. Express y in terms of x. log y = 1.2x - 1 Thank you.
U Unco Senior Member Joined Jul 21, 2005 Messages 1,134 Jan 21, 2006 #2 Don't over-think it . . . If \(\displaystyle \mbox{ y = \log_b{x}}\) \(\displaystyle \mbox{ }\) then \(\displaystyle \mbox{ x = b^y}\) (You are probably expected to assume the log is base 10.)
Don't over-think it . . . If \(\displaystyle \mbox{ y = \log_b{x}}\) \(\displaystyle \mbox{ }\) then \(\displaystyle \mbox{ x = b^y}\) (You are probably expected to assume the log is base 10.)
S soroban Elite Member Joined Jan 28, 2005 Messages 5,584 Jan 21, 2006 #3 Hello, zee! Express \(\displaystyle y\) in terms of \(\displaystyle x:\;\;\log y\;=\;1.2x\,-\,1\) Click to expand... \(\displaystyle \text{We must "exponentiate" both sides:} \L\;e^{^{\ln y}}\;=\;e^{^{1.2x\,-\,1}}\) \(\displaystyle \text{Since }\L e^{^{\ln y}}\,=\,y,\,\)\(\displaystyle \text{ we have: }\L\,y\;=\;e^{^{1.2x-1}}\)
Hello, zee! Express \(\displaystyle y\) in terms of \(\displaystyle x:\;\;\log y\;=\;1.2x\,-\,1\) Click to expand... \(\displaystyle \text{We must "exponentiate" both sides:} \L\;e^{^{\ln y}}\;=\;e^{^{1.2x\,-\,1}}\) \(\displaystyle \text{Since }\L e^{^{\ln y}}\,=\,y,\,\)\(\displaystyle \text{ we have: }\L\,y\;=\;e^{^{1.2x-1}}\)
