Question

I've been having a touch of trouble with solving logarithms with uneven bases. For example 8^x-10=3^x+2
Now I know that in order to solve I need to
log8^x-10=log3^x+2
then you move the power out in front of the logs like so
x-10 log8=x+2 log3
this is where I stumble, you find the approximate value of the log soo
(x-10)(.903)=(x-2)(4.77)
It's after this step were I get lost.

Answer 1

if it was (x-10)*2 = (x-2)*3
what would you do? I would distribute, and combine "like terms"

Answer 2

btw, I assume you meant log(3) = 0.47712 (not 4.77)

Answer 3

so do this
\[ (x-10)(.903)=(x-2)(0.477) \\
0.903x -9.03 = 0.477x -0.954
\]

Answer 4

can you finish?

Answer 5

Yea one sec

Answer 6

.426x=-.051
-/051/.426=-.119

Answer 7

Except x is supposed to = 23.44 ... So I did something wrong

Answer 8

ok, then step by step
\[ 8^{(x-10)}=3^{(x+2)} \\ \log\left( 8^{(x-10)}\right) = \log\left( 3^{(x+2)} \right) \]
\[ (x-10) \log(8) = (x+2) \log(3) \]
rather than approximate the logs, treat them as numbers (which is what they are)

\[ x \log(8) - 10\log(8) = x \log(3) + 2 \log(3) \\
x \log(8) - x \log(3) = 10\log(8) + 2 \log(3) \\
x ( \log(8) - \log(3) ) = 10\log(8) + 2 \log(3) \\
\]
and finally
\[ x = \frac{10\log(8) + 2 \log(3) }{ \log(8) - \log(3)} \]

Answer 9

btw, my first equation I managed to write (x-2) instead of x+2. That was bad.

but, do you follow the steps up above?
to get the final answer you need a calculator

Answer 10

I get 10.57.... not 23.44

Answer 11

what do you get for the top? (btw, I'm getting 23.44102, so it does work out)

Answer 12

I type in 10log(8)+2log(3) and get 9.985...

Answer 13

ok, now the bottom

Answer 14

.4259...

Answer 15

I get 0.425969 which rounds to 0.4260 to 4 digits
but ok
now
9.985/ 0.426

Answer 16

if you keep more digits , you will get a more accurate answer

Answer 17

if you type the expression into google like this
(10 log 8 + 2 log 3)/(log 8 - log 3)=
you get out
23.441021894

Answer 18

but I would learn how to use a calculator (it is a useful to know how to do complicated expressions)