Question

k

Answer 1

I'd start by taking logs of both sides:
\[\Large \log 17(1.14)^x=\log 19(1.16)^x\]
\[\Large \log 17 + \log (1.14)^x=\log 19 + \log (1.16)^x\]
\[\Large \log 17 + x\log 1.14=\log 19 + x\log 1.16\]
Now you can solve.

Answer 2

\[\Large x\log 1.14 -x\log 1.16 =\log 19 -\log 17 \] Notice you have a common factor of x on the left which you can pull out...

Answer 3

The equation is :
\[17\times1.14^x=19\times1.16^x\]
So, by taking the natural logarithm of the two sides :
\[\ln\left(17\times1.14^x\right)=\ln\left(19\times1.16^x\right)\qquad \qquad(1)\]
Now we use the rule :
\[\ln ab=\ln a+\ln b\]
Then (1) becomes :
\[\ln 17+\ln1.14^x=\ln 19+\ln 1.16^x\]
So :
\[x\ln 1.14-x\ln1.16=\ln19-\ln 17 \]
So :
\[x\left(\ln 1.14-\ln1.16\right)=\ln19-\ln17\]
SO :
\[x=\frac{\ln19-\ln17}{\ln 1.14-\ln1.16}\]