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2 years ago

8

Solve the equation 2xlog3=x-1log7

1 answer:

Darya [45]2 years ago

3 0

2xlog3=x-1log7Simplifies to<span><span>0.954243x=x-0.845098</span></span>Let's solve your equation step by step<span><span>0.954243x</span>=<span>x−0.845098</span></span>Step 1:subtract x from both sides.<span><span><span>0.954243x</span>−x</span>=<span><span>x−0.845098</span>−x</span></span><span><span>−<span>0.045757x</span></span>=<span>−0.845098</span></span>Step 2: Divide both sides by -0.045757.<span><span><span>−<span>0.045757x</span></span><span>−0.045757</span></span>=<span><span>−0.845098</span><span>−0.045757</span></span></span><span>x=<span>18.469262</span></span>

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Please prove this........

Crazy boy [7]

Answer: see proof below

<u>Step-by-step explanation:</u>

Given: A + B + C = π → C = π - (A + B)

→ sin C = sin(π - (A + B)) cos C = sin(π - (A + B))

→ sin C = sin (A + B) cos C = - cos(A + B)

Use the following Sum to Product Identity:

sin A + sin B = 2 cos[(A + B)/2] · sin [(A - B)/2]

cos A + cos B = 2 cos[(A + B)/2] · cos [(A - B)/2]

Use the following Double Angle Identity:

sin 2A = 2 sin A · cos A

<u>Proof LHS → RHS</u>

LHS: (sin 2A + sin 2B) + sin 2C

$\text{Sum to Product:}\qquad 2\sin\bigg(\dfrac{2A+2B}{2}\bigg)\cdot \cos \bigg(\dfrac{2A - 2B}{2}\bigg)-\sin 2C$

$\text{Double Angle:}\qquad 2\sin\bigg(\dfrac{2A+2B}{2}\bigg)\cdot \cos \bigg(\dfrac{2A - 2B}{2}\bigg)-2\sin C\cdot \cos C$

$\text{Simplify:}\qquad \qquad 2\sin (A + B)\cdot \cos (A - B)-2\sin C\cdot \cos C$

$\text{Given:}\qquad \qquad \quad 2\sin C\cdot \cos (A - B)+2\sin C\cdot \cos (A+B)$

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$\text{Simplify:}\qquad \qquad 4\cos A\cdot \cos B \cdot \sin C$

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