Solved on Nov 24, 2023

1. a. Solve $2^{x+1}=5^{2x-1}$ for $x$ , exact or rounded to 3 decimals. b. Solve $\log(x^2-10)+\log(9)=1$ for $x$ , give exact (g) and check (s). c. The amount $A$ of a radioactive substance decays over time $t$ (years) as $A(t)=A_0\cdot(1/2)^{t/k}$ , where $k$ is the half-life. Given $A(3)=96.2\%A_0$ , compute the half-life rounded to full years.

STEP 1

Assumptions1. For part a, we are solving the equation $^{x+1}=5^{x-1}$ for $x$ . . For part b, we are solving the equation $\log \left(x^{}-10\right)+\log (9)=1$ for $x$ .
3. For part c, we are given the decay formula $A(t)=A_{0} \cdot\left(\frac{1}{}\right)^{\frac{t}{k}}$ and the information that after3 years, the substance has reduced to96.% of its initial amount. We are asked to compute the half-life of the substance.

STEP 2

For part a, we first take the logarithm base2 on both sides of the equation.
$\log22^{x+1} = \log25^{2x-1}$

STEP 3

We then use the property of logarithms $log_b a^n = n \cdot log_b a$ to simplify the equation.
$x+1 = (2x-1) \cdot \log25$

STEP 4

We solve the equation for $x$ by isolating $x$ on one side.
$x = \frac{1 + \log2}{1 -2 \log2}$

STEP 5

For part b, we first use the property of logarithms $\log a + \log b = \log (ab)$ to simplify the equation.
$\log \left((x^{2}-10) \cdot9\right) =1$

STEP 6

We then use the property of logarithms $\log_b a = c$ is equivalent to $b^c = a$ to rewrite the equation.
$(x^{2}-10) \cdot9 =10$

STEP 7

We solve the equation for $x^2$ by isolating $x^2$ on one side.
$x^2 = \frac{10}{9} +10$

STEP 8

Finally, we take the square root of both sides to solve for $x$ .
$x = \sqrt{\frac{10}{} +10}$

STEP 9

For part c, we first plug in the given values into the decay formula.
$.962A_{} = A_{} \cdot\left(\frac{}{2}\right)^{\frac{3}{k}}$

STEP 10

We then divide both sides by $A_{0}$ to cancel out $A_{0}$ .
$0.962 = \left(\frac{}{2}\right)^{\frac{3}{k}}$

STEP 11

We take the logarithm base0.5 on both sides of the equation.
$\log_{0.5}0.962 = \frac{3}{k}$

STEP 12

Finally, we solve the equation for $k$ by isolating $k$ on one side.
$k = \frac{}{\log_{0.5}0.962}$ The solutions area. $x = \frac{ + \log25}{ -2 \log25}$ b. $x = \sqrt{\frac{10}{9} +10}$ c. $k = \frac{}{\log_{0.5}0.962}$

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STEP 1

STEP 2

For part a, we first take the logarithm base2 on both sides of the equation.
$\log22^{x+1} = \log25^{2x-1}$

STEP 3

We then use the property of logarithms $log_b a^n = n \cdot log_b a$ to simplify the equation.
$x+1 = (2x-1) \cdot \log25$

STEP 4

We solve the equation for $x$ by isolating $x$ on one side.
$x = \frac{1 + \log2}{1 -2 \log2}$

STEP 5

For part b, we first use the property of logarithms $\log a + \log b = \log (ab)$ to simplify the equation.
$\log \left((x^{2}-10) \cdot9\right) =1$

STEP 6

We then use the property of logarithms $\log_b a = c$ is equivalent to $b^c = a$ to rewrite the equation.
$(x^{2}-10) \cdot9 =10$

STEP 7

We solve the equation for $x^2$ by isolating $x^2$ on one side.
$x^2 = \frac{10}{9} +10$

STEP 8

Finally, we take the square root of both sides to solve for $x$ .
$x = \sqrt{\frac{10}{} +10}$

STEP 9

For part c, we first plug in the given values into the decay formula.
$.962A_{} = A_{} \cdot\left(\frac{}{2}\right)^{\frac{3}{k}}$

STEP 10

We then divide both sides by $A_{0}$ to cancel out $A_{0}$ .
$0.962 = \left(\frac{}{2}\right)^{\frac{3}{k}}$

STEP 11

We take the logarithm base0.5 on both sides of the equation.
$\log_{0.5}0.962 = \frac{3}{k}$

STEP 12

Finally, we solve the equation for $k$ by isolating $k$ on one side.
$k = \frac{}{\log_{0.5}0.962}$ The solutions area. $x = \frac{ + \log25}{ -2 \log25}$ b. $x = \sqrt{\frac{10}{9} +10}$ c. $k = \frac{}{\log_{0.5}0.962}$

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STEP 1

STEP 2

For part a, we first take the logarithm base2 on both sides of the equation.log⁡22x+1=log⁡252x−1\log22^{x+1} = \log25^{2x-1}log22x+1=log252x−1

STEP 3

We then use the property of logarithms logban=n⋅logbalog_b a^n = n \cdot log_b alogb​an=n⋅logb​a to simplify the equation.x+1=(2x−1)⋅log⁡25x+1 = (2x-1) \cdot \log25x+1=(2x−1)⋅log25

STEP 4

We solve the equation for xxx by isolating xxx on one side.x=1+log⁡21−2log⁡2x = \frac{1 + \log2}{1 -2 \log2}x=1−2log21+log2​

STEP 5

For part b, we first use the property of logarithms log⁡a+log⁡b=log⁡(ab)\log a + \log b = \log (ab)loga+logb=log(ab) to simplify the equation.log⁡((x2−10)⋅9)=1\log \left((x^{2}-10) \cdot9\right) =1log((x2−10)⋅9)=1

STEP 6

We then use the property of logarithms log⁡ba=c\log_b a = clogb​a=c is equivalent to bc=ab^c = abc=a to rewrite the equation.(x2−10)⋅9=10(x^{2}-10) \cdot9 =10(x2−10)⋅9=10

STEP 7

We solve the equation for x2x^2x2 by isolating x2x^2x2 on one side.x2=109+10x^2 = \frac{10}{9} +10x2=910​+10

STEP 8

Finally, we take the square root of both sides to solve for xxx.x=10+10x = \sqrt{\frac{10}{} +10}x=10​+10​

STEP 9

For part c, we first plug in the given values into the decay formula..962A=A⋅(2)3k.962A_{} = A_{} \cdot\left(\frac{}{2}\right)^{\frac{3}{k}}.962A​=A​⋅(2​)k3​

STEP 10

We then divide both sides by A0A_{0}A0​ to cancel out A0A_{0}A0​.0.962=(2)3k0.962 = \left(\frac{}{2}\right)^{\frac{3}{k}}0.962=(2​)k3​

STEP 11

We take the logarithm base0.5 on both sides of the equation.log⁡0.50.962=3k\log_{0.5}0.962 = \frac{3}{k}log0.5​0.962=k3​

STEP 12

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For part a, we first take the logarithm base2 on both sides of the equation.
$\log22^{x+1} = \log25^{2x-1}$

We then use the property of logarithms $log_b a^n = n \cdot log_b a$ to simplify the equation.
$x+1 = (2x-1) \cdot \log25$

We solve the equation for $x$ by isolating $x$ on one side.
$x = \frac{1 + \log2}{1 -2 \log2}$

For part b, we first use the property of logarithms $\log a + \log b = \log (ab)$ to simplify the equation.
$\log \left((x^{2}-10) \cdot9\right) =1$

We then use the property of logarithms $\log_b a = c$ is equivalent to $b^c = a$ to rewrite the equation.
$(x^{2}-10) \cdot9 =10$

We solve the equation for $x^2$ by isolating $x^2$ on one side.
$x^2 = \frac{10}{9} +10$

Finally, we take the square root of both sides to solve for $x$ .
$x = \sqrt{\frac{10}{} +10}$

For part c, we first plug in the given values into the decay formula.
$.962A_{} = A_{} \cdot\left(\frac{}{2}\right)^{\frac{3}{k}}$

We then divide both sides by $A_{0}$ to cancel out $A_{0}$ .
$0.962 = \left(\frac{}{2}\right)^{\frac{3}{k}}$

We take the logarithm base0.5 on both sides of the equation.
$\log_{0.5}0.962 = \frac{3}{k}$