Question 395226: Solve:
a) log(base 7)(17y + 15) = 2 + log(base 7)(2y - 3)
b) lg(7x-3) + 2 lg 5 = 2 + lg(x + 3)
*Please answer as soon as possible :) :(
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! Solve:
a) log(base 7)(17y + 15) = 2 + log(base 7)(2y - 3)
b) lg(7x-3) + 2 lg 5 = 2 + lg(x + 3)
first problem:
log(base 7)(17y + 15) = 2 + log(base 7)(2y - 3)
log7(17y+15)=2+log7(2y-3)
log7(17y+15)-log7(2y-3)=2
using division rule for logs,
log7(17y+15)/(2y-3)=2
remembering that the base raised to the logarithm of the number is equal to the number. In this case, the base is 7, the logarithm is 2, and the number is (17y+15)/(2y-3)
7^2=(17y+15)/(2y-3)
49=(17y+15)/(2y-3)
98y-147=17y+15
98y-17y=147+15
81y=162
y=2
solving second problem
lg(7x-3) + 2 lg 5 = 2 + lg(x + 3)
using multiplication, power, and division rules of logarithms,
log(7x-3)+log 5^2-log(x+3)=2
log(7x-3)(5^2)/(x+3)=2
using the same rule as above problem
10^2=(7x-3)(5^2)/(x+3)
100=(7x-3)(25)/(x+3)
100=(175x-75)/(x+3)
100x+300=175x-75
75x=375
x=5
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