One More Exponent Question: Hard?

Question

Show me step by step how to solve for 10^2x+3(10)^x -10=0. No matter what I try I get a logarithm with an x or a ^x on one side of the equation. thank you!

Answer 1

Let u = 10^x. You now have u^2 + 3u - 10 = 0

[since 10^2x = (10^x)^2]

That's (u+5)(u-2) = 0
\
So u = -5 or u = 2

Then 10^x = -5 or 10^x = 2

The first is impossible; the second is solved by taking log of both sides:
log(10^x) = log 2; x = log 2

Answer 2

say u=10^x
so in your equation replace 10^x with u's to get
u^2+3u-10=0 (now this a regular quadratic equation, and it is easily factorable... i hope haha)
this factors to (u+5)(u-2)=0
so u=-5 or u=2 (or you could just use the quadratic equation)

first test u=-5
u=10^x
-5=10^x, but you know that 10 to any power can never be negative so:
u cannot be -5

now test u=2
2=10^x, yes this works, log2=x
so the answer is x=log2

i hope that was "step by step" enough for you :)

Answer 3

10^2x+3(10)^x -10=0

let m be 10^x
thus equation will be;

m^2 + 3m - 10 = 0

factors: (m+5)(m-2)=0
m=-5, m=2

m=10^x right?
equate this to the values we have solved:

10^x=-5 (not accepted log negative numbers are invalid)

10^x=2
x=log2/log 10

x=0.301

Answer 4

Hello

Let's factor and we have

(10^x -2)(10^x+5) = 0
so 10^x = 2 or x = log 2 about x = .3010 and 10^x = -5 which is an extraneous root.

Hope This Helps!

Answer 5

Factor out 10^2x:
10^2x(1+3)-10=0
10^2x(4)=10
10^2x=10/4
10^x=sqrt{10} /2
x=log (sqrt10 /2)
x=log(sqrt10)-log2
x=1/2-log2