question on log expression

**janice91** · August 20th 2011, 02:33 AM

Dear all experts,

Would you all please kindly enlighten me with the following equation:

3(power of)2x+5 = 28(power of)x+1

after having a few goes at this equation i have become stuck would any be able to show me the best way to approach it?

Thank you so much for all your help.

**e^(i*pi)** · August 20th 2011, 02:44 AM

Use your exponent laws to split the exponent addition into multiplication: $a^{b+c} = a^ba^c$

$3^{2x} \cdot 3^5 = 28^x \cdot 28^1$

We also know that $a^{bc} = (a^b)^c$ so we can say that $3^{2x} = (3^2)^x = 9^x$

Collect terms with x on the LHS and those without on the RHS: $\dfrac{9^x}{28^x} = \dfrac{28}{3^5}$

We also know that $\dfrac{a^x}{b^x} = \left(\dfrac{a}{b}\right)^x$ and hence the LHS is equal to $\left(\dfrac{9}{28}\right)^x$

Overall therefore: $\left(\dfrac{9}{28}\right)^x = \dfrac{28}{3^5}$

Can you solve for x using logarithms from here?

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Edit: An alternative method is to use factorisation.

Take the log of both sides directly:

$(2x+5)\ln(3) = (x+1)\ln(28)$

Expand the brackets: $2x\ln(3) + 5\ln(3) = x\ln(28) + \ln(28)$

Collect like terms: $2x\ln(3) - x\ln(28) = \ln(28) - 5\ln(3)$

Factor the LHS: $x(2\ln(3) - \ln(28)) = \ln(28)-5\ln(3)$

Divide through to solve for x

Both methods are good so use your favourite choice. They will probably be expressed differently but they're both the same answer $2\ln(3) = \ln(3^2) = \ln(9)$

**Wilmer** · August 20th 2011, 07:22 AM

Janice, tattoo this on your wrist: if a^x = b, then x = log(b) / log(a)

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