Solve the equation. Express the answer in terms of natural logarithms. 3e^9 + t

Question

Solve the equation. Express the answer in terms of natural logarithms. 3e^9 + t = 2 How do oyu go about this? I keep getting the wrong answer... also this one as well: Suppose b is a positive constant greater than 1, and let A, B, and C be defined as follows. logb 2 = A logb 5 = B logb 11 = C In each case, use the properties of logarithms to evaluate the given expression in terms of A, B, and/or C. (Use the change-of-base formula.) (a) log5b 2 b) log5b 55 for a I got a-b and b I got b-b-c and this one (b) log10(x2 ? 36) ? 9[log10(x + 6) + 9 log10(x)] solve for a coefficient of 1 Solve the equation. Give two forms for each answer: one involving base 10 logarithms and the other a calculator approximation rounded to three decimal places. (Enter your answers as comma-separated lists.) 7x2 ? 8 = 12 does anyone have any good techniques for log/ln?

Explanation / Answer

your question is not clear to understand, please take a look at the following answers.

3e^(9+t) = 2

or

e^(9+t) = 2/3

or

9 + t = ln(2/3)

or

t = ln(2/3) - 9

given

logb 2 = A logb 5 = B logb 11 = C

we can have

logb 2 = C

logb 5 = C/A

logb 11 = C/B

(a) log5b 2 = log (2)/log(5b) (note: logy (x) = log10(x)/log10(y))

= log (2)/[log(5)+log(b)] (note: log(xy) = log(x) + log(y))

here take log(b) as common from denominator

= log(2)/[log(b)[log(5)/log(b) + 1]]

= logb (2)/[logb (5) + 1]

= C/[C/A + 1]

= CA/[C + A]

b) log5b 55 = log (5*11)/[log (5b)]

= [log (5) + log (11)]/[log (5) + log (b)]

here take log(b) as common from denominator

= [log (5) + log (11)]/log (b)[log (5)/log (b) + 1]

= [log (5)/log (b) + log (11)/log (b)]/[log (5)/log (b) + 1]

= [logb 5 + logb 11]/[logb 5 + 1]

= [C/A + C/B]/[C/A + 1]

= [CB + CA]/[CB + AB]

(b)

log10(x^2 ? 36) ? 9[log10(x + 6) + log10(x)]

= log10 (x^2 - 36) - log10 ((x+6)x)^9

= log10 [(x + 6)(x - 6)/((x+6)x)^9]

7x^2 ? 8 = 12

or

x^2 = 20/7

or

x = sqrt(20/7) = 1.690

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