Question

math problem

Answer 1

@snowflake0531

Answer 2

@florisalreadytaken

Answer 3

Solution will be a little ugly, but firstly we want to bring variable \(x\) back to base. Any idea?

Hint: https://www.rapidtables.com/math/algebra/logarithm/Logarithm_Rules.html#power%20rule

Answer 4

ok, i told u am busy right -- thats why am replying late
were given:
\[5^{x-3}=7^x\]
lets turn that into a log expression
\[ x-3 = \log_5(7) \ x \]
\[ \ \ \ \ \ \ \ \Updownarrow \]
\[ x= \log_5(7) \ x +3 \]
3 is the constant, so we want to move the expression to the left side(opposite operator):
\[ x-\log_5(7) \ x =3 \]
any clue what were doing next?

Answer 5

sorry i have no clue

Answer 6

you remember we did something similar to this a while ago -- just that the numbers are different -- factor x out

how would that be?

Answer 7

factor x out?

Answer 8

yes
\[ \color{lightskyblue}{x}-\log_5(7) \ \color{lightskyblue}x =3 ​\]
\[ \ \ \ \ \ \ \ \ \ \Updownarrow \]
\[ (1-log_5(7))x = 3 \]
from here its easier -- whats our next step?

Answer 9

factor out 7?

Answer 10

what? no -- that makes no sense

as the previous person said, we have to leave x on its own, as its the base.

to do that we have to divide both sides by \( 1-log_5(7) \) (we tool the paranthesis off as we dont need them anymore)
so that means:
\[ x= \frac{3}{1-log_5(7) } \]

Answer 11

from here we cannot do anything, except using a calculator -- find the nuber, and round it to its nearest TENTH.

Answer 12

-14.3

Answer 13

yeah that looks right

Answer 14

are you sure because i didnt round it

Answer 15

what do you mean?

Answer 16

i didnt round it

Answer 17

show me your calculations please

Answer 18

-14.3 correctly rounded .... what do u mean by "i didnt round it"?

Answer 19

i didnt round it i just rote -14.3

Answer 20

right, do u know what rounding is?

Answer 21

yes