maths : logarithms !! (1 Viewer)

[pLuvia]

There must be an easier way of doing that question but here's my version:
3^2x.5^x=7^3x-1
ln[3^2x.5^x]=ln[7^3x-1]
xln9+xln5=ln7^3x-1
x(ln9+ln5)=ln7^3x-1
x(ln45)=ln7^3x.7^-1
x(ln45)=xln7³ln7^-1
x(ln45-ln343)=-ln7
x=[-ln7]/[ln(45/343)]

 

[ice ken]

Soul Searcher, mind having a go at this one, half the year 12 students cant do this one and yes i go to a comprehensive high school


3^2x*5^x=7^3x-1

we dont need it thats y

 

[z600]

There must be an easier way of doing that question but here's my version:
3^2x.5^x=7^3x-1
ln[3^2x.5^x]=ln[7^3x-1]
xln9+xln5=ln7^3x-1
x(ln9+ln5)=ln7^3x-1
x(ln45)=ln7^3x.7^-1
x(ln45)=xln7³ln7^-1
x(ln45-ln343)=-ln7
x=[-ln7]/[ln(45/343)]

ye thats the long way around