SOLUTION: Log questions? Thank you!
a.) 5^x+1 = 7^3x-2
b.) log2 (x+1) + log2 (x-2) = 2
(where the 2's are small 2's under log)
c.) log5 (x+1) - log5 (x-2) = log5 (x+3)
(
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Question 1009520: Log questions? Thank you!
a.) 5^x+1 = 7^3x-2
b.) log2 (x+1) + log2 (x-2) = 2
(where the 2's are small 2's under log)
c.) log5 (x+1) - log5 (x-2) = log5 (x+3)
(where the 5's are small 5's under log)
Answer by addingup(3677) (Show Source): You can put this solution on YOUR website!
log_2(x+1) + log_2(x-2)= 2 rewrite:
{log(x-2)/log_2}+{log(x+1)/log_2}= 2 Combine fractions:
{(log(x-2))+(log(x+1))/log_2}= 2 Multiply both sides by log_2:
log(x-2)+log(x+1)= 2log_2 Combine on left and simplify on right:
log((x-2)(x+1))= log_4 Take exponents on both sides to get rid of logs:
(x-2)(x+1)= 4 Write the equations on left in standard quadratic form:
x^2-x-2= 4 Add 2 on both sides:
x^2-x= 6 Add 1/4 to both sides (1/4 is one-half of the coefficient of x squared)
x^2-x+1/4= 25/4 Factor:
(x-1/2)^2= 25/4 Take the square root of both sides:
x-1/2= 5/2 Split into 2 equations:
x-1/2= 5/2 or x-1/2= -5/2
x= 3 or x= -2
Test the solutions by substituting for x in the original equations and you will find that x=3 is your answer.
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