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Solve for x.

$11^{-8x} = 18^{-x+10}$

Write the exact answer using base-10 logarithms.

Every online tutorial I've looked up does not have a step by step process explaining how they got it.

  • 1
    is it $ 11^{-8x} = 18^{-x}+10 $ or $11^{-8x} = 18^{-x+10}$? – 2017-01-20
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    It is the one furthest right. – 2017-01-20

2 Answers 2

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$$11^{-8x}=18^{-x+10}$$

Take reciprocals of both sides: $$11^{8x}=18^{x-10}$$

Take the logarithm of both sides and use the identity $\log(a^{b})=b\log(a)$: $$8\log(11)x=\log(18)(x-10)$$

Expand out terms of the right hand side: $$8\log(11)x=\log(18)x-10\log(18)$$

Subtract $x\log(18)$ from both sides: $$(8\log(11)-\log(18))x=-10\log(18)$$

Divide both sides by $8\log(11)-\log(18)$:

$$x=- \frac{10\log(18)}{8\log(11)-\log(18)}\:.$$

  • 0
    Why taking the reciprocal? – 2017-01-20
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    @egreg To get rid of the negative x. – 2017-01-21
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Hint:

$$\frac1{11^{8x}}=\frac{18}{18^x}\implies -8x\log11=\log18-x\log18\implies\ldots$$

Further hint: the above is true for any base of logarithm, not only base $\;10\;$ .