Ask a New Question

Question

Solve log2(x-1) = 5-log2(x+3) for x.

base 2.
13 years ago

Answers

Henry
Log2(x-1) = 5 - Log2(x+3).
Log2(x-1) + Log2(x+3) = 5
Log2((x-1)(x+3)) = 5
Log2(x^2+3x-x-3) = 5
Log2(x^2+2x-3) = 5
x^2 + 2x -3 = 2^5 = 32
x^2 + 2x -3 -32 = 0
x^2 + 2x -35 = 0
(x-5)(x+7) = 0

x-5 = 0
X = 5.

x+7 = 0
X = -7.

Solution set: X = -7, X = 5.
13 years ago

Related Questions

Solve log2(7x-3)>= log2(x+12) My work: 6x=15 x=15/6 x-5/2 so my answer would be: x>= 5/2... Solve: log2(2x^2)[log2(16x)]= 9/2log2x log2^24 - log2^3 = log5^x what is X? thanks solve log2(3x-1)-log2(x-1)=log2(x+1) i have absolutely no idea how to solve this. can anyone help m... 1. The sequence log2 32, log2 y, log2 128, ... forms an arithmetic sequence. What is the value of... Log2(x^4)+log2(x)=5 Find x Simplify log2(8)+log2(2) into single logarithm Convert log2(x)+log2(y) into single logarithm expression 10011 log2 x 11 log2 Simplify 5 log2 k − 8 log2 m + 10 log2 n. 7 log2 (k − m + n) 7 log2 kn over m log2 50 kn ove...
Ask a New Question
Archives Contact Us Privacy Policy Terms of Use