Simplify log⁡x9×log⁡27x\log_x 9 \times \log_{27} x without using a calculator. Solve ln⁡(2x+e)=1+1log⁡xe\ln(2x+e) = 1 + \dfrac{1}{\log_x e}.

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Worked solution

5(a)

1. Apply the change of base formula to convert both logarithms to a common base (e.g., natural logarithm).

log⁡x9×log⁡27x=ln⁡9ln⁡x×ln⁡xln⁡27\log_x 9 \times \log_{27} x = \frac{\ln 9}{\ln x} \times \frac{\ln x}{\ln 27}

2. Cancel the common term $\ln x$ and express 9 and 27 as powers of 3.

=ln⁡9ln⁡27= \frac{\ln 9}{\ln 27}
=ln⁡(32)ln⁡(33)= \frac{\ln (3^2)}{\ln (3^3)}

3. Use the power law of logarithms to simplify and obtain the final numerical value.

=2ln⁡33ln⁡3= \frac{2\ln 3}{3\ln 3}
=23= \frac{2}{3}

Answer

23\frac{2}{3}

5(b)

1. Apply the change of base formula to simplify the fraction

1log⁡xe=log⁡ex=ln⁡x\frac{1}{\log_x e} = \log_e x = \ln x

The equation becomes:
ln⁡(2x+e)=1+ln⁡x\ln(2x+e) = 1 + \ln x

2. Express the right hand side as a single logarithm

Since 1=ln⁡e1 = \ln e,
ln⁡(2x+e)=ln⁡e+ln⁡x\ln(2x+e) = \ln e + \ln x
ln⁡(2x+e)=ln⁡(ex)\ln(2x+e) = \ln(ex)

3. Equate the arguments of the logarithms

2x+e=ex2x + e = ex

4. Solve the linear equation for x

ex−2x=eex - 2x = e
x(e−2)=ex(e - 2) = e
x=ee−2x = \frac{e}{e - 2}

Answer

x=ee−2x = \frac{e}{e - 2}

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