Question

solve algebraically (without using logarithms):

(1/8)^(2x) = 16^(x-5)

Answer 1

\(\color{#000000 }{ \displaystyle \left( 1/8\right)^{2x}=16 ^{x-5}}\)

\(\color{#000000 }{ \displaystyle \left( 1/2^3\right)^{2x}=16 ^{x-5}}\)

\(\color{#000000 }{ \displaystyle \left( 2^{-3}\right)^{2x}=16 ^{x-5}}\)

\(\color{#000000 }{ \displaystyle \left( 2\right)^{(2x)\times (-3)}=16 ^{x-5}}\)

Answer 2

go from there

Answer 3

(oh, i forgot to do the same to the other side, lol)

Answer 4

no, go ahead, if you were typing.
I didn't mean to interupt

Answer 5

\(16=2^4\).
Then, the rule is\((a^b)^c=a^{b\times c}\)

Answer 6

k...

Answer 7

that wasn't me typing by the way

Answer 8

Example problem,

\(\color{#000000 }{ \displaystyle \left(\frac{1}{9} \right)^{3x+5}=81^{x} }\)

We know that \(9=3^2\).

\(\color{#000000 }{ \displaystyle \left(\frac{1}{3^2} \right)^{3x+5}=81^{x} }\)

the rule is: \(\cfrac{1}{a^b}=a^{-b}\)

\(\color{#000000 }{ \displaystyle \left(3^{-2}\right)^{3x+5}=81^{x} }\)

[another] rule is: \((a^b)^c=a^{b\times c}\)

\(\color{#000000 }{ \displaystyle \left(3\right)^{(3x+5)\times (-2)}=81^{x} }\)

simplifying the exponent on the left,

\(\color{#000000 }{ \displaystyle \left(3\right)^{-6x+10}=81^{x} }\)

now, we know that \(81=3^4\)

\(\color{#000000 }{ \displaystyle \left(3\right)^{-6x+10}=(3^4)^{x} }\)

will once again apply: \((a^b)^c=a^{b\times c}\)

\(\color{#000000 }{ \displaystyle \left(3\right)^{-6x+10}=(3)^{(x)\times (4)} }\)

\(\color{#000000 }{ \displaystyle \left(3\right)^{-6x+10}=(3)^{4x} }\)

and we infer that

\(\color{#000000 }{ \displaystyle -6x+10=4x\quad \Rightarrow \quad 10=10x\quad \Rightarrow \quad 1=x }\)

Answer 9

(Reply to your latest comment: I know)

Answer 10

alright just finished, I understand now!

Answer 11

Fabulous!

Answer 12

Another example problem,

\(\color{#000000 }{ \displaystyle \left(\sqrt{5}\right)^{2x+8}=\left(\sqrt{625}\right)^x }\)

\(\color{#000000 }{ \displaystyle \left(5^{1/2}\right)^{2x+8}=\left(625^{1/2}\right)^x }\)

\(\color{#000000 }{ \displaystyle \left(5\right)^{(2x+8)\times (1/2)}=\left(125\right)^{(x)\times (1/2)} }\)

\(\color{#000000 }{ \displaystyle \left(5\right)^{x+4}=\left(625\right)^{\frac{1}{2}x} }\)

\(\color{#000000 }{ \displaystyle \left(5\right)^{x+4}=\left(5^4\right)^{\frac{1}{2}x} }\)

\(\color{#000000 }{ \displaystyle \left(5\right)^{x+4}=\left(5^4\right)^{\left(\frac{1}{2}x \right)\times (4)} }\)

\(\color{#000000 }{ \displaystyle 5^{x+4}=5^{2x } }\)

\(\color{#000000 }{ \displaystyle x+4=2x }\)

\(\color{#000000 }{ \displaystyle 4=x }\)

Answer 13

the idea here is to be able to catch any perfect Nth's when neccesary, and to know the rules of exponents.