Question

can someone help me solve for x please? :)

Answer 1

Where's the question? :)

Answer 2

sorry! there is it :) @Babynini

Answer 3

apply the power rule \[\huge\rm log y^\color{ReD}{x} = \color{reD}{x }\log y\]
`or` you can rewrite 16 in terms of 4 base

Answer 4

i'd just add the log in there or?

Answer 5

hmmm what is say \(\bf 4^2=?\)

Answer 6

16

Answer 7

yes to move the exponent u should take log
here is an example \[\huge\rm 3 ^2 = 2 \log 3\]
but the other way is easier than this one
just rewrite 16 as 4^2 (which is equal to 16)
and then move the base at the numerator \[\large\rm \frac{ 1 }{ x^m }=x^{-m}\]

Answer 8

so xlog4?

Answer 9

4^-2

Answer 10

\[\large\rm 4^x= (\frac{ 1 }{ \color{ReD}{16} })^{2x+5}\]
\[\large\rm 4^x= (\frac{ 1 }{ \color{red}{4^2} })^{2x+5}\]

Answer 11

that's right

Answer 12

ahemm \(\large {
4^x=\left( \cfrac{1}{{\color{brown}{ 16}}} \right)^{2x+5}\implies 4^x=\left( \cfrac{1}{{\color{brown}{ 4^2}}} \right)^{2x+5}\implies 4^x=\left( {\color{brown}{ 4^{-2}}} \right)^{2x+5}
}\)

Answer 13

\[\large\rm 4^x= (4^{-2})^{2x+5}\]
exponent rule \[\rm (a^m)^n = a^{m \times n}\]
distribute parentheses by -2

Answer 14

do i distribute the -2 to the 2x+5?

Answer 15

oh ok

Answer 16

so 4^-4x-10

Answer 17

\[\huge\rm 4^x=4^{(-4x-10)}\]
bases are the same so you can cancel them
\[\huge\rm \cancel{4}^x=\cancel{4}^{(-4x-10)}\]
x=-4x -10 solve for x that's it

Answer 18

ohh thanks!!

Answer 19

yw

Answer 20

could you help me on another one please?

Answer 21

hmm sorry im abt to hit logout :(

Answer 22

ok thanks anyways :)