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2 years ago

9

What is x for this problem? Please help!

1 answer:

meriva2 years ago

6 0

The answer for x is 3

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2log(x-3)+1=5<br><br> can someone explain how to solve for x?

Brums [2.3K]

$\bf 2log(x-3)+1=5\iff 2log_{10}(x-3)+1=5\\\\\\ 2log_{10}(x-3)=4
\\\\\\
log_{10}(x-3)=\cfrac{4}{2}\implies log_{10}(x-3)=2\\\\
-----------------------------\\\\
{{ a}}^{log_{{ a}}x}=x\impliedby \textit{using this cancellation rule}\\\\
-----------------------------\\\\
10^{\cfrac{}{}log_{10}(x-3)}=10^2\implies x-3=10^2\implies x=100+3$

5 0

3 years ago

W+3 3/8 =1 5/6=<br> I know the answer but i want to see if you get it

nasty-shy [4]

Answer:

...

Step-by-step explanation:

7 0

2 years ago

Read 2 more answers

Solve for y.<br> 22mm+15mm+(4y+12)mm=180 degrees

AURORKA [14]

Answer:

y = 45/m^2 − 49/4

Isolate the variable by dividing each side by factors that don't contain the variable

3 0

3 years ago

Y = 6x - 4<br> y = -x + 3

likoan [24]

Answer:

(1, 2)

Step-by-step explanation:

Subtract the second equation from the first:

(y) -(y) = (6x -4) -(-x +3)

0 = 7x -7 . . . . . simplify

0 = x - 1 . . . . . . divide by 7

1 = x . . . . . . . . . add 1

y = -1 +3 = 2 . . . substitute into the second equation

The solution is (x, y) = (1, 2).

4 0

3 years ago

A mixture of compounds X and Y in a 0.100-cm cell had an absorbance of 0.215 at 272 nm and 0.191 at 327 nm. Find [X] and [Y] in

Oksi-84 [34.3K]

Answer: The concentration of X is $7.13582\times 10^{-6}$ and the concentration of Y is $2.53159\times 10^{-5}$ .

Step-by-step explanation:

Since we have given that

At 272 nm, absorbance = 0.215

At 327 nm, absorbance = 0.191

As we have given that

Compound X Compound Y

272 16400 3870

327 3990 6420

So, our equations becomes

$16400C_1+3870c_2=0.215\\\\3990C_1+6420C_2=0.191$

By solving these two equations, we get that

$C_1=7.13582\times 10^{-6}\\\\C_2=2.53159\times 10^{-5}$

Hence, the concentration of X is $7.13582\times 10^{-6}$ and the concentration of Y is $2.53159\times 10^{-5}$ .

5 0

3 years ago