Can These Two Complex Math Equations Be Solved Correctly?

[RaYaMe]

Hi every body , I had a lot of equations to solve yesterday

becouse that i was preparing my self for my exam this sunday

and I've did .. but these TWO equations .. i couldn't be sure about my results

can you help me to solve them ..

1- The variables p and q are related by the law q=ap^b , where a and b are constants . Given that ln(p) = 1.32 when ln(q)= 1.73 and ln(p)=0.44 when ln(q)=1.95 ,
find the values of a and b .

2- Write logb (x) + n as one logarithm​

* b is the base

​

 

[tiny-tim]

Welcome to PF!

Hi RaYaMe! Welcome to PF!

(try using the X² tag just above the Reply box )

Show us what you got, and then we'll know how to help!

 

[RaYaMe]

for the second one ..
2- Write logb (x) + n as one logarithm

i've tried =>

logb (x) + logb (b)^n = logb (X.b^n) ?? or it's canceled



AND ..the first one .. i tried to take ln of both sides ( q=ap^b)

lnq = lna + b lnp

It'll be >> 1- 1.73= lna + 1.32 b

>> 2- 1.95 = lna + 0.44 b

that's what I've did to get the value of " b "

1.73=lna + 1.32 b
-1.95 = -lna - 0.44 b
___________________
-0.22 = 0.88 b >>>>> b= -0.22/0.88 = -0.25

1.73 = lna + ( 1.32 * -0.25 )
1.73 + 0.33 = lna
2.06 = lna

so , a = e^2.06 what do you think about my ways ..

 

[tiny-tim]

logb (x) + logb (b)^n = logb (X.b^n)
…
-0.22 = 0.88 b >>>>> b= -0.22/0.88 = -0.25

1.73 = lna + ( 1.32 * -0.25 )
1.73 + 0.33 = lna
2.06 = lna

so , a = e^2.06 what do you think about my ways ..

Yes, that looks fine!