Question

Solve for x & show all steps:
2^x = 64
5^2x = 5^10

Now, I obviously know the answers to these but my teacher marks hard as hell so I need to know the steps behind it because she assumes I'm guessing.

Answer 1

i dont think theres really steps behind the first one ;-;

Answer 2

she's a horrid marker and im sure there aren't even steps behind it

Answer 3

For the Second, Since they have an equivalent base of 5, you just solve the exponents. so cancel out the 5's and solve 2x=10. divide by 2 and get x=10.

Answer 4

x=5* o-o Omg.

Answer 5

haha

Answer 6

ok, got it. i'll apply this to the test tomorrow and if she marks it like she has been im just gonna confront her lol

Answer 7

Okay xD GOODLUCK c:

Answer 8

make both sides have the *same* base
then you can equate the exponents

Answer 9

ok so what if the bases are different like what if i had 5^3x = 7^10 or something, how would i go about doing that

Answer 10

sorry about all these random questions, school just started and it's not fresh in my mind yet

Answer 11

for the 2nd problem, you already are set
\[ 5^{2x} = 5^{10} \\ 2x= 10 \\ x= 5 \]

Answer 12

if you have different bases, then you use a more "sophisticated" idea. take the log of each side.
\[ 5^{3x} = 7^{10} \\ \log\left(5^{3x} \right)= \log\left(7^{10} \right) \\
3x \log(5) = 10 \log(7) \\
x = \frac{ 10 \log(7)}{3 \log(5)}
\]

Answer 13

for the first problem
2^x = 64

notice 64 is a power of 2. (If you did not know, you could hope it's true, and divide by 2 to get 2*32, and then 2*2*16, and then 2*2*2*8, and then 2*2*2*2*2*2 = 2^6

Answer 14

oh okay, that seems good to do

Answer 15

it might be slightly trickier if you see
\[ 9^x = 27^2 \]
27 is not a power of 9. the next power 9^2 = 9*9=81
*but*, notice we could factor 9 into 3*3= 3^2 and 27 is 3*3*3 = 3^3
and we have
\[ \left(3^2\right)^x = \left( 3^3\right)^2 \]
use this property (good to know)
\[ \left(a^b\right)^c = a^{bc} \]
to write
\[ 3^{2x}= 3^{3 \cdot 2}\\ 3^{2x}= 3^6\]
and 2x=6 , and x= 6/2 = 3

Answer 16

2^x = 64