Question

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Answer 1

a) looks like a simple linear equation, is that correct?

Answer 2

Correction on a. Its actually 3^(2x) = 5(3^x) + 36. i forgot to add the ^ symbols

Answer 3

ahh thought so

Answer 4

Sorry about that:)

Answer 5

for the first, try rewriting in quadratic form:
(3^x)^2 -5(3^x) -36 = 0

Answer 6

ok you can either make everything have same base, then set exponents equal
or like in part a) make substitution u = 3^x, it becomes a quadratic

Answer 7

u^2 - 5u -36 = 0
(u-9)(u+4) = 0
u=9, -4
so now solve:
3^x = 9 and 3^x = -4 but you already know this second one does not have a solution.

Answer 8

All of these are exponential equations. You'll need to use logarithms to solve. Hopefully that will help.

Answer 9

For (a)
3^(2x) = 5(3^x) + 36
(3^x)^(2) = 5(3^x) + 36
(3^x)^(2) - 5(3^x) - 36 =0
[(3^x) -9][(3^x)+4] =0
[(3^x) -9]=0 or [(3^x)+4] =0
(3^x) = 9 or (3^x)=-4
xlog3 = log9 or xlog3 = log(-4) (rejected)
x=2

Answer 10

@dpalnc 3^(2x) =/= (3^x)^2

Answer 11

\[1/8^{x-3}=2\times16^{2x-1}\]
=>
\[2^{-3(x-3)} = 2\times2^{4(2x+1)}\]
=>
\[2^{-3x+9}=2^{(8x+4)+1}\]
=>
Since both sides are equal then we can say that their exponents are equal, thus we get:
\[-3x+9 = 8x+5\]
Solve for X.

Answer 12

I only did the one problem. You can use the same technique to solve for the others.

Answer 13

Also, as with pretty much every math problem, make sure you check your answers. :)

Answer 14

technique needed for (c) is similar to that in (b) so the answer from Quaternions can serve as a reference for you

Answer 15

Thank you so much for helping me guys! @clallisto, i particularly found your information very helpful! Il try to do my best with this questions. I really appreciate the time and help for everyone who answered! Have a blessed day! :)

Answer 16

welcome. God blessed! try (c) now ~