PreCalc: Simplifying Logs?

Question

Sorry, lots of questions again! I'm such a bonehead at maths!!

(a) How do I write 2InX + 3In((x^0.5)y) as a single logarithm

(b) How do I solve for x: 2In (3x+6) = 6

(c) How do I solve for x: e^((x-2)^2) = 2

(d) How do I solve for x: 7^(-2x) = 3

I'm quite sure I have c and d correct, so If someone can't be bothered typing the explanation then just the answer is fine! (just to be sure, there's no answers provided for my set applications!)

Thanks everyone!!
Tessa

Thanks so much blitz!

They're all right!! Except B, I got a completely different answer, would you be able to explain your result for B?

Thanks, xx

Answer 1

(a) How do I write 2InX + 3In((x^0.5)y) as a single logarithm

ln x^2 + ln ((x^0.5)y)^3
ln [x^2((x^0.5)y)^3]

(b) How do I solve for x: 2In (3x+6) = 6

ln (3x + 6)^2 = 6
take e on both sides
(3x + 6)^2 = e^6
square root both sides
3x + 6 = +- e^3
3x = -6 +- e^3
x = (1/3)(-6 +- e^3)

(c) How do I solve for x: e^((x-2)^2) = 2

take ln on both sides
(x - 2)^2 = ln 2
(x - 2) = (ln 2)^(1/2)
x = 2 + (ln 2)^(1/2)

(d) How do I solve for x: 7^(-2x) = 3

take ln on both sides
(-2x) ln 7 = ln 3
-2x = ln 3 / ln 7
x = (-1/2)(ln 3 / ln 7)

Answer 2

a) 2InX + 3In((x^0.5)y)
=InX^2 + In((x^0.5)y)^3
=ln(X^2*((x^0.5)y)^3)
=ln (X^(7/2)*Y^3)

b)2In (3x+6) = 6
In (3x+6) = 3
3x+6=e^3
x=(e^3-6)/3

c)e^((x-2)^2) = 2
(x-2)^2 * lne =ln2

x-2= + or - sqrt(ln2)

x=+ or - sqrt(ln2)+2

d) 7^(-2x) = 3
-2x ln 7=ln3

x= - ln3/2ln7

Answer 3

A.
2InX + 3In((x^0.5)y) = ln(X^2[(X^0.5)Y)^3)]
= ln ((x^3.5)(Y^3))
B.
(3x+6)^2 = e^6
expand and solve the x value

C.

(X-2)^2 = ln 2

D.

-2x = lg3 / lg7

Answer 4

assets of log : log(ab) = loga + logb additionally,loga + logb = logab further log(a/b) = loga - logb different properties are : loga^ok = kloga log (base b^ok) a = (a million/ok) log(base b)a u ought to locate a solid e book.

Answer 5

sure as what u dont understand