Find, to 3 significant figures, the value of x for which 5^x=7.

Hence, or otherwise, solve the equation 5^2x -12(5^x)+ 35=0

 Solution

log⁡〖5^x 〗=log⁡7

xlog⁡5=log⁡7

×=log⁡7/log⁡5

×=1.20906…

×=1.21(3 sf).

5^2x-12(5^x )+35=0

Solution

(5^x )^2-12(5^x)+35=0

Let y=5^x

y^2-12y+35=0

(y-5)  (y-7)=0

Either y-5=0 or y-7=0

y=5 or y=7

5^x=5 or 5^x=7

×=1 or ×=1.21(3sf).