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Chapter 3: Problem 111

Solve the logarithmic equation algebraically. Approximate the result to three decimal places. $$\log 8 x-\log (1+\sqrt{x})=2$$

Short Answer

Expert verified
The solution to the given logarithmic equation is \(x = 10.000\)

Step by step solution

01

Use the properties of logarithms

We want to simplify the equation by applying the properties of logarithms. In this case, we can apply the quotient rule for logarithms which is \(\log b(a) - \log b(c) = \log b(\frac{a}{c})\). Therefore, the given equation, \(\log 8 x-\log (1+\sqrt{x}) = 2\), will become \(\log ( \frac{8x}{1 + \sqrt{x}}) = 2\)
02

Remove the logarithm

To remove the logarithm, we will use the property of logarithms that \(b^{\log_b x} = x\). This gives us the reciprocal property of \(\log_b {b^x} = x\). By applying this to the equation from step 1, we get \(\frac{8x}{1 + \sqrt{x}} = 10^2\).
03

Simplify the equation

The equation from step 2 simplifies to \(8x = 100 + 10 \cdot \sqrt{x}\). Rearrange it to get \(8x - 10\cdot\sqrt{x} - 100 = 0\). This gives a quadratic equation in terms of \(\sqrt{x}\) which is \(8(\sqrt{x})^2 - 10\sqrt{x} - 100 = 0 \). Thereafter, use the quadratic formula \(\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) to solve for \(\sqrt{x}\)
04

Solve the quadratic equation

Solving the quadratic equation from step 3 gives \(\sqrt{x} = \frac{10 \pm \sqrt{100 + 4*8*100}}{2*8}\) . We ignore the negative root since it results in an imaginary number. The positive root gives \(\sqrt{x} = 3.16228 \). Squaring both sides to get the equation in the form of \(x\), we get \(x = 3.16228^2\)
05

Approximate the solution

After solving from step 4, we get \(x = 3.16228^2\), hence \(x = 10\). To approximate to three decimal places, we get \(x = 10.000\). Therefore the solution of given logarithmic equation is \(x = 10.000\)

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Most popular questions from this chapter

If $$\$ 1$$ is invested in an account over a 10-year period, the amount in the account, where \(t\) represents the time in years, is given by \(A=1+0.075 \llbracket t \rrbracket\) or \(A=e^{0.07 t}\) depending on whether the account pays simple interest at \(7 \frac{1}{2} \%\) or continuous compound interest at \(7 \%\). Graph each function on the same set of axes. Which grows at a higher rate? (Remember that \(\llbracket t \rrbracket\) is the greatest integer function discussed in Section 1.6.)

Use the acidity model given by \(\mathrm{pH}=-\log \left[\mathrm{H}^{+}\right],\) where acidity \((\mathrm{pH})\) is a measure of the hydrogen ion concentration \(\left[\mathrm{H}^{+}\right]\) (measured in moles of hydrogen per liter) of a solution. Compute \(\left[\mathrm{H}^{+}\right]\) for a solution in which \(\mathrm{pH}=3.2\).

Determine the time necessary for $$\$ 1000$$to double if it is invested at interest rate \(r\) compounded (a) annually, (b) monthly, (c) daily, and (d) continuously. $$r=6.5 \%$$

Solve the logarithmic equation algebraically. Approximate the result to three decimal places. $$\ln (x+5)=\ln (x-1)-\ln (x+1)$$

COMPARING MODELS If $$\$ 1$$ is invested in an account over a 10 -year period, the amount in the account, where \(t\) represents the time in years, is given by \(A=1+0.06 \llbracket t \rrbracket\) or \(A=[1+(0.055 / 365)]^{[365 t]}\) depending on whether the account pays simple interest at \(6 \%\) or compound interest at \(5 \frac{1}{2} \%\) compounded daily. Use a graphing utility to graph each function in the same viewing window. Which grows at a higher rate?
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