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.
$\endgroup$ 22 Answers
$\begingroup$$$11^{-8x}=18^{-x+10}$$
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 $8\log(11)-\log(18)$:
$$x=- \frac{10\log(18)}{8\log(11)-\log(18)}\:.$$
$\endgroup$ 2 $\begingroup$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 $\;10\;$ .
$\endgroup$
