Solve the given exponential equation?

Question

The equation is:

5^x = 25^x-1

I know you have to make 25^x-1 and put it as (5^2)^x-1

Answer 1

do u mean 5^x = 25^ (x-1) or 5^x = 25^x -1. you need to express clearly.

if 5^x = 25^(x -1)

5^x = 5^(2x-2)
take log on both sides,

log 5^x = log 5^ (2x-2)

x log5 = 2x-2 log5, { ust log a^b= b loga }

we leave out log 5, then ,we get

x = 2x -2

2 = x,
x =2

Answer 2

Q: Is it 5^x = 25^ (x-1) OR 5^x = 25^x -1 ?

If it's 5^x = 25^(x-1), you simply express 25 as 25 = 5^2.

by Law of Exponents which says (u^a)^b = u^(ab),

5^x = 5^[2(x-1)]

recall: if u^a = u^b , then a = b.

Thus, x = 2(x-1).

Solving for x, x = 2.

let's check it out: if x = 2, 5^x = 5^2 = 25.

25^(x-1) = 25^(2-1) = 25^1 = 25.

Clearly, 5^x = 25^(x-1) when x = 2.

Answer 3

logarithm the two section xln19=ln143 x=ln143/ln19 use your calculator 2. 2^x=z z^2+z-12=0 (z+4)(z-3)=0 z= -4 z=3 2^x= -4 no answer 2^x=3 x=ln3/ln2 use your calculator 3. x-7=5^2 x=25+7=32 4. x^3=one hundred twenty 5 x=5 5. 5( x-2)=one hundred x-2=20 x=22

Answer 4

let a=5^x
so 5^x=25^x-1 becomes a=a^2-1
or a^2-a-1=0 the positive solution is a=(1+sqr(5))/2 = 5^x

so taking logs gives x=0.298994

Answer 5

5^x = 25^(x - 1)
5^x = (5^2)^(x - 1)
5^x = 5^(2(x - 1))
5^x = 5^(2x - 2)

therefore

x = 2x - 2
-x = -2
x = 2

ANS : x = 2

Answer 6

5^x = 25^(x-1)
5^x = (5^2)^(x-1)
5^x = 5^(2(x-1))
5^x = 5^(2x-2)
x = 2x-2
x = 2

Answer 7

If, on the other hand, folks, you take the equation EXACTLY as written (and ammended) thus making it

5^x = 25^x - 1

the answer will come out approximately 0.2516 by iteration, since there is no closed form solution Im aware of........

Answer 8

x = 2

5^2= 25
25 ^(2-1) = 25

Answer 9

5^x = 25^x-1
5^x = 5^2(x-1)
x=2x-1
-1x= -1
x=1