solve the expression 2log 5-1/2log 16+2log 40

asked by guest
on Apr 10, 2025 at 2:05 am



You asked:

Evaluate the expression: 2log⁡(5)−12log⁡(16)+2log⁡(40)2 \log\left( 5 \right) - \frac{1}{2 \log\left( 16 \right)} + 2 \log\left( 40 \right)

MathBot Answer:

2log⁡(5)−12log⁡(16)+2log⁡(40)=−12log⁡(16)+2log⁡(5)+2log⁡(40)≈10.416297852984952928986439476528572 \log\left( 5 \right) - \frac{1}{2 \log\left( 16 \right)} + 2 \log\left( 40 \right) = - \frac{1}{2 \log{\left(16 \right)}} + 2 \log{\left(5 \right)} + 2 \log{\left(40 \right)} \approx 10.41629785298495292898643947652857

2log⁡(5)−12log⁡(16)+2log⁡(40)=2⋅log⁡(5)−12⋅log⁡(16)+2⋅log⁡(40)=2⋅log⁡(5)−12⋅1log⁡(16)+2⋅log⁡(40)=(2⋅log⁡(5)−12⋅1log⁡(16))+2⋅log⁡(40)=2⋅log⁡(5)+2⋅log⁡(40)−12⋅1log⁡(16)\begin{aligned}2 \log\left( 5 \right) - \frac{1}{2 \log\left( 16 \right)} + 2 \log\left( 40 \right)&=2 \cdot \log{\left(5 \right)} - \frac{1}{2 \cdot \log{\left(16 \right)}} + 2 \cdot \log{\left(40 \right)}\\&=2 \cdot \log{\left(5 \right)} - \frac{1}{2} \cdot \frac{1}{\log{\left(16 \right)}} + 2 \cdot \log{\left(40 \right)}\\&=\left(2 \cdot \log{\left(5 \right)} - \frac{1}{2} \cdot \frac{1}{\log{\left(16 \right)}}\right) + 2 \cdot \log{\left(40 \right)}\\&=2 \cdot \log{\left(5 \right)} + 2 \cdot \log{\left(40 \right)} - \frac{1}{2} \cdot \frac{1}{\log{\left(16 \right)}}\end{aligned}

+ Submit Question