Question

3^(5x+2)=7^(x-1)

How/why is ln used to solve this type of equations??

Answer 1

Because if you remember, in an equation, in order to mantain the equality, you have to put the inverse of what you got on the other side, a product goes to the other side as a division, a plus changes to a minus and so on. There you have an exponential function so the inverse of the exponential function is the Ln. That's why you use it to solve.

Answer 2

y = log ax, a^y = x

Answer 3

it has to do with a change of base procedure

Answer 4

3^(5x+2)=7^(x-1)
Ln(3^5x+2)= ln(7^x-1)
(5x+2) ln(3)= (x-1) ln(7)
But that as far as I can go.. Idk what to do next? Do I distribute orrrr divide etc.?

Answer 5

let ln(7)/ln(3) = k

therefore

5x+2 = (x-1) * k

5x+2 = kx - k

5x-kx = -2 - k

x(5-k) = -2 - k

x = -(2+k)/(5-k)

Answer 6

another way to see that was to go this route:
\[3^{5x+2}=7^{x-1}\]
\[L_3(7^{x-1})=5x+2\]
\[(x-1)L_3(7)=5x+2\]
\[L_3(7)=\frac{L(7)}{L(3)}=k\]
\[(x-1)k=5x+2\]

Answer 7

My study guide says the answer is x= -2 log(3) - log(7)/ 5 log(3) - log(7)

Answer 8

it most likely simplifies to that after alot of heartache :)

Answer 9

Ugh, too much heartache lol

Answer 10

\[x = -\frac{(2+k)}{(5-k)}\]
\[x = -\frac{(2+\frac73)}{(5-\frac73)}\]
\[x = -\frac{(2.3+.7)}{(5-\frac73).3}\]
\[x = -\frac{(2.3+.7)\cancel{.3}}{(5.3-.7)\cancel{.3}}\]
\[x = \frac{-2.3-.7}{5.3-.7}\]

Answer 11

How do you differentiate when to use log(x) or ln(x)

Answer 12

recall that the derivative of log(x) is 1/x

say: y = 3x

lny = ln(3x) , derive

y'/y = 3/(3x) , solve for y'

y' = y/x
= 3x/x
=3

Answer 13

logx = lnx/ln10
derive = 1/xln(10)