logs problem

[Guest]

I have this question that i cant solve, i woked through it but i cant find what i did wrong. please help?

If 5(23^x)=47^(1-x) solve for X using base ten

Log5+xlog23=(1-x)Log47
Log5=(1-x)Log47-xLog23
Log5=x((1-1)Log47-1Log23)
X=Log5+Log23

but thats wrong. what am i doing wrong :?

 

[galactus]

I have this question that i cant solve, i woked through it but i cant find what i did wrong. please help?

If 5(23^x)=47^(1-x) solve for X using base ten

Log5+xlog23=(1-x)Log47
Log5=(1-x)Log47-xLog23
Log5=x((1-1)Log47-1Log23)
X=Log5+Log23

but thats wrong. what am i doing wrong :?

\(\displaystyle \L\\5\cdot\23^{x}=47^{1-x}\)

\(\displaystyle \L\\log(5\cdot\23^{x})=log(47^{1-x})\)

\(\displaystyle \L\\log(5)+log(23^{x})=log(47^{1-x})\)

\(\displaystyle \L\\log(5)+xlog(23)=(1-x)log(47)\)

\(\displaystyle \L\\log(5)+xlog(23)=log(47)-xlog(47)\)

\(\displaystyle \L\\xlog(23)+xlog(47)=log(47)-log(5)\)

\(\displaystyle \L\\x=\frac{log(47)-log(5)}{log(23)+log(47)}\)

\(\displaystyle \H\\=\frac{log(\frac{47}{5})}{log(1081)}\)

 

[Guest]

Thank you ....i forgot to distribute the brackets.

thank you very much