Question

Question regarding logarithms.

Answer 1

x = 6

Explanation:

5^(x - 2) = 5^(x - 3) × 5¹

= 5[5^(x - 3)]

5^(x - 4) = 5^(x - 3) ÷ 5¹

= [5^(x - 3)]/5

7^(x - 5) = 7^(x - 3) ÷ 7²

= [7^(x - 3)]/49

5[5^(x - 3)] - 7^(x - 3) = [7^(x - 3)]/49 + 11[5^(x - 3)]/5

5[5^(x - 3)] - 11[5^(x - 3)]/5 = [7^(x - 3)]/49 + 7^(x - 3)

5^(x - 3)[5 - 11/5] = 7^(x - 3) [1/49 + 1]

5^(x - 3)[14/5] = 7^(x - 3) [50/49]

7^(x - 3) ÷ 5^(x - 3) = 14/5 × 49/50

(7/5)^(x - 3) = 7³/5³ = (7/5)³

x - 3 = 3

x = 6

Answer 2

`5^(x-2)-7^(x-3)=7^(x-5)+11\cdot5^(x-4)\\\\5^(x-2)-7^(x-2-1)=7^(x-2-3)+11\cdot5^(x-2-2)\qquad\text{use}\ (a^n)/(a^m)=a^(n-m)\\\\5^(x-2)-(7^(x-2))/(7^1)=(7^(x-2))/(7^3)+11\cdot(5^(x-2))/(5^2)\\\\5^(x-2)-(1)/(7)\cdot7^(x-2)=(1)/(343)\cdot7^(x-2)+(11)/(25)\cdot5^(x-2)\\\\-(1)/(7)\cdot7^(x-2)-(1)/(343)\cdot7^(x-2)=(11)/(25)\cdot5^(x-2)-5^(x-2)\\\\\left(-(1)/(7)-(1)/(343)\right)\cdot7^(x-2)=\left((11)/(25)-1\right)\cdot5^(x-2)`

`\left(-(49)/(343)-(1)/(343)\right)\cdot7^(x-2)=-(14)/(25)\cdot5^(x-2)\\\\-(50)/(343)\cdot7^(x-2)=-(14)/(25)\cdot5^(x-2)\qquad\text{multiply both sides by}\ \left(-(25)/(14)\right)\\\\(50\cdot25)/(343\cdot14)\cdot7^(x-2)=5^(x-2)\qquad\text{divide both sides by}\ 7^(x-2)\\\\(25\cdot25)/(343\cdot7)=(5^(x-2))/(7^(x-2))\qquad\text{use}\ \left((a)/(b)\right)^n=(a^n)/(b^n)`

`(5^2\cdot5^2)/(7^3\cdot7)=\left((5)/(7)\right)^(x-2)\qquad\text{use}\ a^n\cdot a^m=a^(n+m)\\\\(5^4)/(7^4)=\left((5)/(7)\right)^(x-2)\\\\\left((5)/(7)\right)^4=\left((5)/(7)\right)^(x-2)\iff x-2=4\qquad\text{add 2 to both sides}\\\\\boxed{x=6}`