15 Evaluate log_(2)3 __ 16 Evaluate log_(2)sqrt (16) __ 17 Given 7^7x-5=7^2x solve for x __ 18 If 9^2x=243^x+3 what is the value of x __ 19 Given log_(2)4=log_(2)9x solve for x __ 20 If log2+logx=log4 , solve for x __ 21 Solve for x: log_(5)(x+1)=log_(5)2 __ 22 Convert 230^circ into exact radians __ 23 Convert 313^0 into approximate radians __ 24 State the exact value of cos((5pi )/(3)) __ 25 State the exact value of sin((5pi )/(4)) __ 26 State the exact value of sec((7pi )/(6)) __ 27 Solve sinTheta =-(1)/(2),0leqslant Theta leqslant 2pi __ 28 Solve tanTheta =-1,0leqslant Theta leqslant 2pi __ 29 Solve cosTheta =(4)/(5),0leqslant Theta leqslant 2pi __ 30 Solve sinTheta =-(1)/(sqrt (2)),0leqslant Theta leqslant 2pi __ 31 Express sinxcos3x+cosxsin3x as single trig function __ 32 If f(x)=3sin[2(x+(pi )/(6))]-1 state phase and vertical trans __

# Explanation:<br />## Step 1: Evaluate \( \log _{2} 3 \)<br />### This is a logarithmic expression that cannot be simplified further without a calculator. The value of \( \log _{2} 3 \) is approximately 1.58496.<br /># Answer:<br />### \( \log _{2} 3 \approx 1.58496 \)<br /><br /># Explanation:<br />## Step 1: Evaluate \( \log _{2} \sqrt{16} \)<br />### Rewrite the square root as an exponent: \( \sqrt{16} = 16^{\frac{1}{2}} \).<br />## Step 2: Simplify the logarithm<br />### Use the property of logarithms: \( \log_b(a^c) = c \log_b(a) \). Thus, \( \log_2(16^{\frac{1}{2}}) = \frac{1}{2} \log_2(16) \).<br />## Step 3: Calculate \( \log_2(16) \)<br />### Since \( 16 = 2^4 \), we have \( \log_2(16) = 4 \).<br />## Step 4: Final calculation<br />### Therefore, \( \frac{1}{2} \log_2(16) = \frac{1}{2} \times 4 = 2 \).<br /># Answer:<br />### \( \log _{2} \sqrt{16} = 2 \)<br /><br /># Explanation:<br />## Step 1: Given \( 7^{7x-5}=7^{2x} \), solve for \( x \)<br />### Since the bases are the same, set the exponents equal to each other: \( 7x - 5 = 2x \).<br />## Step 2: Solve the equation<br />### Subtract \( 2x \) from both sides: \( 5x - 5 = 0 \).<br />### Add 5 to both sides: \( 5x = 5 \).<br />### Divide by 5: \( x = 1 \).<br /># Answer:<br />### \( x = 1 \)<br /><br /># Explanation:<br />## Step 1: Given \( 9^{2x}=243^{x+3} \), solve for \( x \)<br />### Rewrite the bases as powers of 3: \( 9 = 3^2 \) and \( 243 = 3^5 \).<br />### Thus, \( (3^2)^{2x} = (3^5)^{x+3} \).<br />## Step 2: Simplify the exponents<br />### \( 3^{4x} = 3^{5(x+3)} \).<br />### Set the exponents equal: \( 4x = 5(x + 3) \).<br />## Step 3: Solve the equation<br />### Distribute: \( 4x = 5x + 15 \).<br />### Subtract \( 5x \) from both sides: \( -x = 15 \).<br />### Multiply by -1: \( x = -15 \).<br /># Answer:<br />### \( x = -15 \)<br /><br /># Explanation:<br />## Step 1: Given \( \log _{2} 4=\log _{2} 9x \), solve for \( x \)<br />### Since the logarithms are equal, set the arguments equal: \( 4 = 9x \).<br />## Step 2: Solve the equation<br />### Divide by 9: \( x = \frac{4}{9} \).<br /># Answer:<br />### \( x = \frac{4}{9} \)<br /><br /># Explanation:<br />## Step 1: Given \( \log 2+\log x=\log 4 \), solve for \( x \)<br />### Use the property of logarithms: \( \log a + \log b = \log(ab) \).<br />### Thus, \( \log(2x) = \log 4 \).<br />## Step 2: Set the arguments equal<br />### \( 2x = 4 \).<br />## Step 3: Solve the equation<br />### Divide by 2: \( x = 2 \).<br /># Answer:<br />### \( x = 2 \)<br /><br /># Explanation:<br />## Step 1: Solve for \( x \): \( \log _{5}(x+1)=\log _{5} 2 \)<br />### Since the logarithms are equal, set the arguments equal: \( x + 1 = 2 \).<br />## Step 2: Solve the equation<br />### Subtract 1: \( x = 1 \).<br /># Answer:<br />### \( x = 1 \)<br /><br /># Explanation:<br />## Step 1: Convert \( 230^{\circ} \) into exact radians<br />### Use the conversion factor \( \pi \text{ radians} = 180^{\circ} \).<br />### Thus, \( 230^{\circ} \times \frac{\pi}{180^{\circ}} = \frac{230\pi}{180} \).<br />## Step 2: Simplify the fraction<br />### \( \frac{230\pi}{180} = \frac{23\pi}{18} \).<br /># Answer:<br />### \( 230^{\circ} = \frac{23\pi}{18} \text{ radians} \)<br /><br /># Explanation:<br />## Step 1: Convert \( 313^{\circ} \) into approximate radians<br />### Use the conversion factor \( \pi \text{ radians} = 180^{\circ} \).<br />### Thus, \( 313^{\circ} \times \frac{\pi}{180^{\circ}} \approx 5.46288 \text{ radians} \).<br /># Answer:<br />### \( 313^{\circ} \approx 5.46288 \text{ radians} \)<br /><br /># Explanation:<br />## Step 1: State the exact value of \( \cos \left(\frac{5 \pi}{3}\right) \)<br />### Recognize that \( \frac{5\pi}{3} \) is in the fourth quadrant where cosine is positive.<br />### The reference angle is \( 2\pi - \frac{5\pi}{3} = \frac{\pi}{3} \).<br />### Thus, \( \cos \left(\frac{5 \pi}{3}\right) = \cos \left(\frac{\pi}{3}\right) = \frac{1}{2} \).<br /># Answer:<br />### \( \cos \left(\frac{5 \pi}{3}\right) = \frac{1}{2} \)<br /><br /># Explanation:<br />## Step 1: State the exact value of \( \sin \left(\frac{5 \pi}{4}\right) \)<br />### Recognize that \( \frac{5\pi}{4} \) is in the third quadrant where sine is negative.<br />### The reference angle is \( \frac{5\pi}{4} - \pi = \frac{\pi}{4} \).<br />### Thus, \( \sin \left(\frac{5 \pi}{4}\right) = -\sin \left(\frac{\pi}{4}\right) = -\frac{\sqrt{2}}{2} \).<br /># Answer:<br />### \( \sin \left(\frac{5 \pi}{4}\right) = -\frac{\sqrt{2}}{2} \)<br /><br /># Explanation:<br />## Step 1: State the exact value of \( \sec \left(\frac{7 \pi}{6}\right) \)<br />### Recognize that \( \frac{7\pi}{6} \) is in the third quadrant where cosine is negative.<br />### The reference angle is \( \frac{7\pi}{6} - \pi = \frac{\pi}{6} \).<br />### Thus, \( \cos \left(\frac{7 \pi}{6}\right) = -\cos \left(\frac{\pi}{6}\right) = -\frac{\sqrt{3}}{2} \).<br />### Therefore, \( \sec \left(\frac{7 \pi}{6}\right) = \frac{1}{\cos \left(\frac{7 \pi}{6}\right)} = -\frac{2}{\sqrt{3}} = -\frac{2\sqrt{3}}{3} \).<br /># Answer:<br />### \( \sec \left(\frac{7 \pi}{6}\right) = -\frac{2\sqrt{3}}{3} \)<br /><br /># Explanation:<br />## Step 1: Solve \( \sin \theta=-\frac{1}{2}, 0 \leq \theta \leq 2 \pi \)<br />### Identify the angles where \( \sin \theta = -\frac{1}{2} \).<br />### These angles are in the third and fourth quadrants.<br />### The reference angle is \( \frac{\pi}{6} \).<br />### Thus, \( \theta = \pi + \frac{\pi}{6} = \frac{7\pi}{6} \) and \( \theta = 2\pi - \frac{\pi}{6} = \frac{11\pi}{6} \).<br /># Answer:<br />### \( \theta = \frac{7\pi}{6}, \frac{11\pi}{6} \)<br /><br /># Explanation:<br />## Step 1: Solve \( \tan \theta=-1,0 \leq \theta \leq 2 \pi \)<br />### Identify the angles where \( \tan \theta = -1 \).<br />### These angles are in the second and fourth quadrants.<br />### The reference angle is \( \frac{\pi}{4} \).<br />### Thus, \( \theta = \pi - \frac{\pi}{4} = \frac{3\pi}{4} \) and \( \theta = 2\pi - \frac{\pi}{4} = \frac{7\pi}{4} \).<br /># Answer:<br />### \( \theta = \frac{3\pi}{4}, \frac{7\pi}{4} \)<br /><br /># Explanation:<br />## Step 1: Solve \( \cos \theta=\frac{4}{5}, 0 \leq \theta \leq 2 \pi \)<br />### Identify the angles where \( \cos \theta = \frac{4}{5} \).<br />### These angles are in the first and fourth quadrants.<br />### Using the inverse cosine function: \( \theta = \cos^{-1} \left(\frac{4}{5}\right) \) and \( \theta = 2\pi - \cos^{-1} \left(\frac{4}{5}\right) \).<br /># Answer:<br />### \( \theta = \cos^{-1} \left(\frac{4}{5}\right), 2\pi - \cos^{-1} \left(\frac{4}{5}\right) \)<br /><br /># Explanation:<br />## Step 1: Solve \( \sin \theta=-\frac{1}{\sqrt{2}}, 0 \leq \theta \leq 2 \pi \)<br />### Identify the angles where \( \sin \theta = -\frac{1}{\sqrt{2}} \).<br />### These angles are in the third and fourth quadrants.<br />### The reference angle is \( \frac{\pi}{4} \).<br />### Thus, \( \theta = \pi + \frac{\pi}{4} = \frac{5\pi}{4} \) and \( \theta = 2\pi - \frac{\pi}{4} = \frac{7\pi}{4} \).<br /># Answer:<br />### \( \theta = \frac{5\pi}{4}, \frac{7\pi}{4} \)<br /><br /># Explanation:<br />## Step 1: Express \( \sin x \cos 3 x+\cos x \sin 3 x \) as a single trig function<br />### Use the sum-to-product identities: \( \sin A \cos B + \cos A \sin B = \sin(A + B) \).<br />### Thus, \( \sin x \cos 3x + \cos x \sin 3x = \sin(x + 3x) = \sin(4x) \).<br /># Answer:<br />### \( \sin x \cos 3 x+\cos x \sin 3 x = \sin(4x) \)<br /><br /># Explanation:<br />## Step 1: If \( f(x)=3 \sin \left[2\left(x+\frac{\pi}{6}\right)\right]-1 \), state phase and vertical translation<br />### Identify the phase shift and vertical translation from the function.<br />### The phase shift is given by \( x + \frac{\pi}{6} \), which means a shift of \( -\frac{\pi}{6} \).<br />### The vertical translation is given by the constant term outside the sine function, which is -1.<br /># Answer:<br />### Phase shift: \( -\frac{\pi}{6} \), Vertical translation: -1