SOLUTION: find x 7^x = 4^x+1 that you do this but i dont know how find x log7^x=log4^x+1 xlog7=xlog4+1log4 x(log7-xlog4)=1log4 x= 1log4/log7-log4

Question 171560: find x
7^x = 4^x+1
that you do this but i dont know how find x
log7^x=log4^x+1
xlog7=xlog4+1log4
x(log7-xlog4)=1log4
x= 1log4/log7-log4
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
find x
7^x = 4^(x+1)
Take the log of both sides to get:
xlog(7) = (x+1)log(4)
xlog(7)-xlog(4) = log4
x(log(7/4)) = log4
x = [log(4)]/[log(7/4)]
x = 2.477225...
Cheers,
Stan H.

RELATED QUESTIONS
what is the way to solve the logarithms a.) 2e^9x=4 b.)ln(4x − 3) = 4 and c.)... (answered by lwsshak3)

log7^4x + log7^(x+1) =... (answered by longjonsilver)

Log7^2=x... (answered by josgarithmetic)

log7 X + log7 (x-6)=log7... (answered by edjones)