logarithms (please help)

[zee]

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.

 

[Unco]

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.)

 

[soroban]

Hello, zee!

Express \(\displaystyle y\) in terms of \(\displaystyle x:\;\;\log y\;=\;1.2x\,-\,1\)

\(\displaystyle \text{We must "exponentiate" both sides:} \L\;e^{^{\ln y}}\;=\;e^{^{1.2x\,-\,1}}\)


\(\displaystyle \text{Since }\L e^{^{\ln y}}\,=\,y,\,\)\(\displaystyle \text{ we have: }\L\,y\;=\;e^{^{1.2x-1}}\)