Any Idea how to solve 8(10^-x)+10^x = 6 ?

Question

Any idea or perhaps would you take your time to sovle this equation?

Thanks

Answer 1

Let 10^x = A
Then 8/A + A = 6
8+ A^2 = 6A
A^2 - 6A + 8 = 0
(A -2)(A - 4) = 0
A = 2 or A = 4
10^x = 2 or 10^x = 4
x = log 2 or x = log 4
(Logarithms are base 10)

Answer 2

8(10^-x)+10^x = 6

8(1/10^x) + 10^x = 6

8/(10^x)+10^x = 6

(8+(10^x)^2)/10^x = 6

multiply both sides by 10^x to cancel the divisor
or just cross multiple... next outcome will be this:

8+(10^x)^2 = 6*(10^x)

(10^x)^2 - 6(10^x) +8 = 0

(10^x - 2) * (10^x - 4) = 0

10^x = 2 | 10^x =4

x = log 2 | x = log 4

x = 0.301029 | x = 0.602059

That's all I have for now.... I will look for a less complicated solution when I get the time... Hope it helps....

Answer 3

8(10^(-x)) + 10^x = 6

Change 10^(-x) to 1/10^x.

8(1/10^x) + 10^x = 6

Multiply both sides by 10^x.

8 + (10^x)^2 = 6(10^x)

Move everything to the left hand side.

(10^x)^2 - 6(10^x) + 8 = 0

This is a quadratic in disguise. To show this, let u = 10^x.

u^2 - 6u + 8 = 0

Now, factor

(u - 4)(u - 2) = 0

This implies u = 4, {so 10^x = 4} or u = 2 {so 10^x = 2).

If 10^x = 4, then x = log(4).
If 10^x = 2, then x = log(2).

Therefore, your solutions are x = {log(4), log(2)}

Answer 4

Hi /
That is not hard to solve if u just get orginzed

8(10^-x)+10^x = 6
fist u need to get ride of the (^-x)..by multiplying by 10^X

(8(10^-x)+10^x = 6) *10^X --->
(8(1)+10^(x^2)=6*10^x ---> quadratic equation

10^(x^2)-6*10^x-8=0
U think u can solve it from there?
GL

Answer 5

well really simple:
do this step by step:
1)8*10^(-x)+10^x=6 -> 8/(10^x)+10^x=6 ->muiltiply all in 10^x
so 8+10^(2x)=6*10^x now in step 2)
2)consider 10^x=y so 10^2x=y^2 and ur equation becomes:
8+y^2=6y -> y^2-6y+8=0 it is quadratic formula:
3)the roots are :
6+ sqrt(4))/2 and (6-sqrt(4))/2 so:
y1=4 and y2=2
ok in step 3 we considerd y=10^x so we want x ok put y in the formula:
4=10^x -> x=log4
and
2=10^x ->x=log2
so ur final answers are :
x=log4 and x=log2
logarithms are base 10 ok
goodluck!

Answer 6

8/10^x+10^x=6
8+10^2x-6*10^x=0
set 10^x=t
t^2-6t+8=0
(t-4)(t-2)=0
t=4 or 2
10^x=4 x=log 4 base 10
10^x=2 x=log 2 base 10

Answer 7

Anywhere you see something to the power of x, you should take the logarithm of the equation. Then you can use:

log (a^x) = x log a

to simplify it.

Answer 8

take the log of both sides:
log8+(-x)log10+xlog10=log6
log8 is equal to log2^3 and log2 is equal to 0.3 so log8 is 0.9
and as you know log10 is 1
so >>>>0.9+(-x)+x=log2+log3
and then we have 0.9=log2+log3
so: 0.6=log3
and because it's p<==>q the answer seems to be right.I hope so.......

Answer 9

try taking logarithms...on both sides of the equal to sign....i think that can lead you ahead