Hey Jspjulie,
Lets solve your problem step by step.
The question we are encountering is: How do I solve : [tex]4e^{0.1x}=60[/tex].
The first step is to divide both sides by 4.
[tex]\dfrac{4e^{0.1x}}{4}=\dfrac{60}{4}[/tex]
Now we simplify the fraction
[tex]e^{0.1x}=15[/tex]
[tex]\mathrm{If\:}f\left(x\right)=g\left(x\right)\mathrm{,\:then\:}\ln \left(f\left(x\right)\right)=\ln \left(g\left(x\right)\right)[/tex]
[tex]\ln \left(e^{0.1x}\right)=\ln \left(15\right)[/tex]
Now we have to apply the log rule: [tex]\log _a\left(x^b\right)=b\cdot \log _a\left(x\right)[/tex]
[tex]0.1x\ln \left(e\right)=\ln \left(15\right)[/tex]
Now we simplify
[tex]0.1x=\ln \left(15\right)[/tex]
Finally, we reach our answer
[tex]x=\ln \left(15\right)\cdot \:10[/tex]
Hope this helps,
AnthrαX
