What is the answer to log_{2}(3-x)=log_{4}(2x^{18}) ?

The answer to log_{2}(3-x)=log_{4}(2x^{18}) is x\approx 1.03709…,x\approx-1.12634…

Solve for x: log_{2}(3-x)=log_{4}(2x^{18})

The solution is x\approx 1.03709…,x\approx-1.12634…

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lo g_{2} (3 - x) = lo g_{4} (2 x^{18})

x \approx 1.03709 \dots, x \approx - 1.12634 \dots

Solution steps

lo g_{2} (3 - x) = lo g_{4} (2 x^{18})

Apply log rules

(3 - x)^{2} = 2 x^{18}

Solve

(3 - x)^{2} = 2 x^{18} : x \approx 1.03709 \dots, x \approx - 1.12634 \dots

x \approx 1.03709 \dots, x \approx - 1.12634 \dots

Verify Solutions:

x \approx 1.03709 \dots

True

, x \approx - 1.12634 \dots

True

The solutions are

x \approx 1.03709 \dots, x \approx - 1.12634 \dots

2 lo g_{10} (x) = lo g_{10} (10 - 3 x)

ln (x) + 5 = 10

lo g_{3} (x + 1) - lo g_{3} (27) = 4

lo g_{10} (n) = 2.5955

lo g_{3} (2 x + 1) - lo g_{3} (5 x - 3) = - 1

What is the answer to log_{2}(3-x)=log_{4}(2x^{18}) ?
The answer to log_{2}(3-x)=log_{4}(2x^{18}) is x\approx 1.03709…,x\approx-1.12634…
Solve for x: log_{2}(3-x)=log_{4}(2x^{18})
The solution is x\approx 1.03709…,x\approx-1.12634…