Asked by Jessi Chytka on May 21, 2024
Verified
Solve the logarithmic equation. Round your answer to two decimal places. log2x+log2(x+1) −log23=4\log _ { 2 } x + \log _ { 2 } ( x + 1 ) - \log _ { 2 } 3 = 4log2x+log2(x+1) −log23=4
A) x=2.00x = 2.00x=2.00
B) x=−3.00;x=2.00x = - 3.00 ; x = 2.00x=−3.00;x=2.00
C) x=6.45x = 6.45x=6.45
D) x=−7.45;x=6.45x = - 7.45 ; x = 6.45x=−7.45;x=6.45
E) none of these
Logarithmic Equation
An equation that involves the logarithm of an expression equal to a value or another logarithm.
- Solve logarithmic equations and understand the properties of logarithms.
Verified Answer
ZN
Zeenab NoralMay 24, 2024
Final Answer :
C
Explanation :
Using logarithm rules, we can simplify the equation as follows:
log2(x(x+1)3)=4\log_2\left(\frac{x(x+1)}{3}\right)=4log2(3x(x+1))=4
x(x+1)3=24=16\frac{x(x+1)}{3}=2^4=163x(x+1)=24=16
x2+x−48=0x^2+x-48=0x2+x−48=0
Solving for x, we get $x=6.45$ or $x=-7.45$. However, since we cannot take the logarithm of a negative number, we discard the negative solution. Therefore, the only valid solution is $x=6.45$.
log2(x(x+1)3)=4\log_2\left(\frac{x(x+1)}{3}\right)=4log2(3x(x+1))=4
x(x+1)3=24=16\frac{x(x+1)}{3}=2^4=163x(x+1)=24=16
x2+x−48=0x^2+x-48=0x2+x−48=0
Solving for x, we get $x=6.45$ or $x=-7.45$. However, since we cannot take the logarithm of a negative number, we discard the negative solution. Therefore, the only valid solution is $x=6.45$.
Learning Objectives
- Solve logarithmic equations and understand the properties of logarithms.