How do you solve $8^(6x-5)=(1/16)^(6x-2)$ ?

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Answer 1 · 2016-03-03T04:03:21

x = $23/42$

$a^(m+n) = a^m.a^n$
Express 8 and 16 in powers of 2. cross multiply.
$2^(42x-23)$ = 1.
$a^0=1$ .
$42x-23$ =0..

Answer 2 · 2016-03-03T04:05:25

$x=23/42$

We will write $8$ and $1/16$ as powers of $2:$

$8=2^3$

$1/16=1/2^4=2^-4$

The equation can then be rewritten as

$(2^3)^(6x-5)=(2^-4)^(6x-2)$

We then use the rule:

$(a^b)^c=a^(bc)$

This makes the equation

$2^(3(6x-5))=2^(-4(6x-2))$

Since the bases are equal, the exponents can be set equal to one another as well:

$3(6x-5)=-4(6x-2)$

$18x-15=-24x+8$

$42x=23$

$x=23/42$

How do you solve 8^(6x-5)=(1/16)^(6x-2) 8^(6x-5)=(1/16)^(6x-2)?