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Not if you are capable of operating a scientific calculator.

I read your equation as: 1/2* log(5x) + log(5) = 1

Solve for x:
log(5) + 1/2 log(5 x) = 1

Rewrite the left hand side by combining fractions. log(5) + 1/2 log(5 x) = 1/2 (2 log(5) + log(5 x)):
1/2 (2 log(5) + log(5 x)) = 1

Multiply both sides by 2:
2 log(5) + log(5 x) = 2

Subtract 2 log(5) from both sides:
log(5 x) = 2 - 2 log(5)

2 - 2 log(5) = 2 + log(1/5^2) = 2 + log(1/25):
log(5 x) = 2 + log(1/25)

Cancel logarithms by taking exp of both sides:
5 x = e^(2 + log(1/25))

e^(2 + log(1/25)) = e^(log(1/25)) e^2 = e^2/25:
5 x = (e^2)/(25)

Divide both sides by 5:
Answer: | x = e^2/125

(I think you meant to put cm² )

area = 54

length = 2 + x

width = 2x

area = (length)(width)    Plug in values given in the problem.

54    =   (2 + x)  (2x)       Distribute

54 = 4x + 2x²                Subtract 54 from both sides and rearrange.

0 = 2x² + 4x - 54           Divide through by 2

0 = x² + 2x - 27             Use the quadraic formula to solve for x.

\(x = {-2 \pm \sqrt{2^2-4(1)(-27)} \over 2(1)}= {-2 \pm 4\sqrt{7} \over 2}=-1\pm 2\sqrt7\)

x ≈ 4.292     or     x ≈ -6.292

If the length is 2 cm more than x, the question probably wants the positive value for x. To the nearest whole numer that is x = 4 cm

(-1.84-1.56)/5 \(\stackrel{?}{=}\) 8

(-3.4)/5 \(\stackrel{?}{=}\) 8

-0.68 ≠ 8

You are right, it doesn't equal 8!

Maybe there is more to the question? What do the instructions say?

Hectictar gave a great answer to this only yesterday so go find it!

Hint - half the lengths of the sides are 3,4 and 2

you must find the hypotenuses of these right angled triangles.

Thank you!

It would have been to both our benefits if you had mentioned yesterday that you were looking for real or imaginary factors.

This is a young persons forum so it is inappropriate for me to assume any question asker knows anything about complex numbers!

Factorize (x2 -0.7x+0.49) Using imaginary numbers, into two terms such that:  (x- something) (x- something) ?

\(If\;\;x^2 -0.7x+0.49=0\;\;then\\x = {-b \pm \sqrt{b^2-4ac} \over 2a}\\ x = {--0.7 \pm \sqrt{0.49-4*0.49} \over 2}\\ x = {0.7 \pm \sqrt{-3*0.49} \over 2}\\ x = {0.7 \pm 0.7 \sqrt{3}i \over 2}\\ x = 0.35 \pm 0.35 \sqrt{3}i \\ x = 0.35(1+\sqrt{3}i )\quad or\quad x = 0.35(1-\sqrt{3}i )\\\\~\\ so\\ x^2 -0.7x+0.49=(x-0.35(1+\sqrt{3}i )(x+0.35(1+\sqrt{3}i ))\\ x^2 -0.7x+0.49=(x-0.35-0.35\sqrt{3}i )(x+0.35+0.35\sqrt{3}i )\)

WHAT IS A RULE FOR THE SEQUENCE 1.8 , 2.85 , 3.90 , 4.95 , 6

it is just adding 1.05 each time

so a=1.8   first term and d (common difference) = 1.08

\(T_n=a+(n-1)d\\ T_n=1.8+1.05(n-1)\\ T_n=1.8+1.05n-1.05\\ T_n=0.75+1.05n\)

original area = a

a = 200 * 300

2a = (200 + x) * (300 + x)

Substitute in the value we know for a.

2(200 * 300) = (200 + x) * (300 + x)

Solve for x.

120,000 = 60,000 + 500x + x²

60,000 = 500x + x²

0 = x² + 500x - 60,000

0 = (x - 100)(x + 600)

x = 100     or     x = -600

Since the question says we are adding length, x must = 100

\(2\sec^2x+\tan^2x-3=0\\~\\ 2\cdot\frac{1}{\cos^2x}+\frac{\sin^2x}{\cos^2x}-\frac{3\cos^2x}{\cos^2x}=0\\~\\ \frac{2+\sin^2x-3\cos^2x}{\cos^2x}=0\\~\\ 2+\sin^2x-3\cos^2x=0 \qquad , \, x\neq\frac{\pi}{2}+n\pi \\~\\ 2 + (1-\cos^2x)-3\cos^2x=0 \\~\\ 3 - 4\cos^2x=0 \\~\\ \cos^2x=\frac34 \\~\\ \cos x=\frac{\sqrt3}{2}\)

From looking at a unit circle, the angles within [0, 2π) that have a cosine of \( \frac{\sqrt3}{2} \) are \( \frac{\pi}{6} \) and \( \frac{11\pi}{6} \)   

Think of it this way, 9+-6 is the exact same thing as 9-6.

The sequence continues as: 9/5, 57/20, 39/10, 99/20, 6, 141/20, 81/10, 183/20......etc.

Each term can be found as: a(n) = 3/20 (7n + 5) (for all terms given).

Solve for x:
x^2 - 0.7 x + 0.49 = 0

x^2 - 0.7 x + 0.49 = x^2 - (7 x)/10 + 49/100:
x^2 - (7 x)/10 + 49/100 = 0

Subtract 49/100 from both sides:
x^2 - (7 x)/10 = -49/100

Add 49/400 to both sides:
x^2 - (7 x)/10 + 49/400 = -147/400

Write the left hand side as a square:
(x - 7/20)^2 = -147/400

Take the square root of both sides:
x - 7/20 = (7 i sqrt(3))/20 or x - 7/20 = -(7 i sqrt(3))/20

Add 7/20 to both sides:
x = 7/20 + (7 i sqrt(3))/20 or x - 7/20 = -(7 i sqrt(3))/20

Add 7/20 to both sides:
Answer: | x = 7/20 + (7 i sqrt(3))/20       or        x = 7/20 - (7 i sqrt(3))/20

\(\frac{\text{sector area}}{\text{circle area}}=\frac{\text{central angle measure}}{\text{the total angular measure of a circle}} \\~\\ \frac{\text{sector area}}{\pi*4^2}=\frac{\frac{5\pi}2}{2\pi} \\~\\ \text{sector area}=(\pi*4^2)(\frac{\frac{5\pi}2}{2\pi})=20\pi\approx20(3.14)=62.8\text{ sq. ft.}\)