Ayanlaja Adebola Debo

1 year ago

1 year ago

A 19 year old boy named Ayanlaja Ramoni Adebola from Nigeria has just discovered is own mathematical formula for solving mathematical problem of a^x=bx. A question was asked by his friend to him which brought his research to find the solution to the mathematical problem. He was asked to find the value of x in the equation 4^x=8x.He had to make research on such questions hoping they exist. But he discovered that most people know that the answer was 2 but what is the solution(workings) to the problem? Some made use of graph method, some made use of Eisten Raph method but he noticed most people had to limit the value of 2 because they already knew the answer was 2.

He made his research on such question and concluded there must be workings leading to the solution of the problem.

By his handwork he proposed his own formula which he named "Debo's formula". Which states that for the equation a^x=bx, then x={log{b/loga)} /loga. (where x is approximated to a whole number).

Chek out the solution of 4^x=8x using Debo's formula.

4^x=8x

Solution

a=4 and b=8

Debo's formula ==>x=log{b/loga} divided by log a

X=log{8/log4} /log4

X=log{8/0.6021} /log4

X=log 13.2868/log4

X=1.1234/0.6021

X=1.86=2(Approximate to the nearest whole number)

Thus, using Debo's formula x=2.

Proof that x=2 from Debo's formula

4^x=8x

4^2=8*2

16=16

His friends argued that his formula can only be used to solve 4^x=8x...Another question of that type was created.

Solve 3^x=9x

Using Debo's formula

a=3 and b=9

x=log{b/loga} divided by loga

x=log{9/log3) divided by log3

x=log(9/0.4771) /log3

x=log18. 864/log3

x=1.2756/0.4771

x=2.67=3(approximate to nearest whole number)

X=3 using Debo's formula.

Proof ==>

3^x=9x

3^3=9*3

27=27

Using Debo's formula

Another problem was created

8^x=32x

Using Debo's formula

a=8 and b=32

Debo's formula ==>x=log{b/loga} divided by loga

x=log{32/log8} /log8

x=log {32/0.9031}/log 8

X=log 35.43/log8

X=1.5494/0.903

X=1.7=2(Approximate to the nearest whole number)

Using Debo's formula x=2

Proof that x=2

8^x=32x

8^2=32*3

64=64

The conditions in using Debo's formula are both the LHS and RHS numeric value must be both even or both odd. Also the numeric value at exponential part (a^x) must be lesser and be divisible by the numeric value of the side with identity x. Thus in a^x=bx, a

He made his research on such question and concluded there must be workings leading to the solution of the problem.

By his handwork he proposed his own formula which he named "Debo's formula". Which states that for the equation a^x=bx, then x={log{b/loga)} /loga. (where x is approximated to a whole number).

Chek out the solution of 4^x=8x using Debo's formula.

4^x=8x

Solution

a=4 and b=8

Debo's formula ==>x=log{b/loga} divided by log a

X=log{8/log4} /log4

X=log{8/0.6021} /log4

X=log 13.2868/log4

X=1.1234/0.6021

X=1.86=2(Approximate to the nearest whole number)

Thus, using Debo's formula x=2.

Proof that x=2 from Debo's formula

4^x=8x

4^2=8*2

16=16

His friends argued that his formula can only be used to solve 4^x=8x...Another question of that type was created.

Solve 3^x=9x

Using Debo's formula

a=3 and b=9

x=log{b/loga} divided by loga

x=log{9/log3) divided by log3

x=log(9/0.4771) /log3

x=log18. 864/log3

x=1.2756/0.4771

x=2.67=3(approximate to nearest whole number)

X=3 using Debo's formula.

Proof ==>

3^x=9x

3^3=9*3

27=27

Using Debo's formula

Another problem was created

8^x=32x

Using Debo's formula

a=8 and b=32

Debo's formula ==>x=log{b/loga} divided by loga

x=log{32/log8} /log8

x=log {32/0.9031}/log 8

X=log 35.43/log8

X=1.5494/0.903

X=1.7=2(Approximate to the nearest whole number)

Using Debo's formula x=2

Proof that x=2

8^x=32x

8^2=32*3

64=64

The conditions in using Debo's formula are both the LHS and RHS numeric value must be both even or both odd. Also the numeric value at exponential part (a^x) must be lesser and be divisible by the numeric value of the side with identity x. Thus in a^x=bx, a

berber

2 years ago

2 years ago

4^x=8x,

(1 3)^x=8x,

1 3x x(x-1)/2=8x,

1 3x (x^2-x)/2=8x,

(2 6x x^2-x)/2=8x,

2 6x x^2-x=16x,

x^2 6x-x-16x 2=0,

x^2-11x 2=0. Solve the equation quadratically,u will get 2.

(1 3)^x=8x,

1 3x x(x-1)/2=8x,

1 3x (x^2-x)/2=8x,

(2 6x x^2-x)/2=8x,

2 6x x^2-x=16x,

x^2 6x-x-16x 2=0,

x^2-11x 2=0. Solve the equation quadratically,u will get 2.

jiddey100

47 years ago

47 years ago

Please I also need the answer to the above question. Anybody with the correct answer should please help with a solution. Thanks.

millonnism

2 years ago

2 years ago

X=2 if not still contented with cross check by substitute x=2 in the equation.

Igboanugo Okwudili

1 year ago

1 year ago

Thanks Guys bt i have not seen any meaningful proof yet

#Opeyemi

2 years ago

2 years ago

using logbase2.... logbase2 (4^x) =logbase2 (8x) 2xlogbase2 (2)=logbase2 (8)+logbase2 (x) 2x=3logbase2 (2)+logbase2 (x) 2x=3 + logbase2 (x) 2x-3=logbase2 (x) 2^(2x-3)=x^1 2=x and 2x-3=1 [2x-3=1;2x=4;x=2]

Azeez Lateef Tunde

1 year ago

1 year ago

4^x = 8x

Introduce natural logathrims to both sides

Log4^x = log8x

From loga + logb = logab

.: log4^x = log8 + logx

Express the numbers in their simplest form

Log2^2x = log2^3 + logx

apply the rule of power law

2xlog2 = 3log2 + logx

By Rearranging the terms

logx = 2xlog2 - 3log2

by factorizing the expression

Logx = log2(2x - 3)

by dividing through by log2

Logx/log2 = (2x - 3)

Note loga/logb = a/b

Therefore we have,

x/2 = (2x - 3)

Multiply through by 2

x = 4x - 6

By collecting their like term,

-3x = -6

Divide both sides by -3

X = 2 ..........................proved

Introduce natural logathrims to both sides

Log4^x = log8x

From loga + logb = logab

.: log4^x = log8 + logx

Express the numbers in their simplest form

Log2^2x = log2^3 + logx

apply the rule of power law

2xlog2 = 3log2 + logx

By Rearranging the terms

logx = 2xlog2 - 3log2

by factorizing the expression

Logx = log2(2x - 3)

by dividing through by log2

Logx/log2 = (2x - 3)

Note loga/logb = a/b

Therefore we have,

x/2 = (2x - 3)

Multiply through by 2

x = 4x - 6

By collecting their like term,

-3x = -6

Divide both sides by -3

X = 2 ..........................proved

david497

2 years ago

2 years ago

4^x=8x

soln: 4^x=8x

4^x=x(2^2+2^2)

4^x=x(4)^2

4^x=x(2)^2×2

2^2x=2x^4

2x=4

dividing both sides by 2.

x= 2

soln: 4^x=8x

4^x=x(2^2+2^2)

4^x=x(4)^2

4^x=x(2)^2×2

2^2x=2x^4

2x=4

dividing both sides by 2.

x= 2

realmind

1 year ago

1 year ago

This question seems to have no true solution...4^x=8x. And if there is I would like to know it.

Abumgod

2 years ago

2 years ago

4'x = 8x -----eq1

2'2x = (2³)x

x = 2'2x/2³

x = 2'(2x-3)

sub x in eq1

4'(2'(2x-3)) = 8(2'(2x-3))

16'(2x-3) = 2³(2'(2x-3))

2'(4(2x-3)) = 2'(3+2x-3)

base cancels out

4(2x-3) = 2x

8x - 12 = 2x

6x =12

x = 2 = Final answer!

2'2x = (2³)x

x = 2'2x/2³

x = 2'(2x-3)

sub x in eq1

4'(2'(2x-3)) = 8(2'(2x-3))

16'(2x-3) = 2³(2'(2x-3))

2'(4(2x-3)) = 2'(3+2x-3)

base cancels out

4(2x-3) = 2x

8x - 12 = 2x

6x =12

x = 2 = Final answer!

Mahmmudlawal

1 year ago

1 year ago

4^x=8x

Divide through by x

Then, 4^x/x =8

4^x/x =16/2 since 16/2 =8

Equate either the numerator or denominator

4^x =16 =4^2

X=2

Proved.......

Divide through by x

Then, 4^x/x =8

4^x/x =16/2 since 16/2 =8

Equate either the numerator or denominator

4^x =16 =4^2

X=2

Proved.......

Hugo15

1 year ago

1 year ago

Hey guys, to solve that you'll have know a particular rule thats,

LogaN=logbN/logbA.

So, 4'x=8x

XLog(base 2)4=3log(base 2)2 + log(base 2)x. (you'll get that after simplifying)

then,

2x=3 +log(base10)x/log(base10)2

2x=3+x/2

4x=6+x

3x=6, x=2.

LogaN=logbN/logbA.

So, 4'x=8x

XLog(base 2)4=3log(base 2)2 + log(base 2)x. (you'll get that after simplifying)

then,

2x=3 +log(base10)x/log(base10)2

2x=3+x/2

4x=6+x

3x=6, x=2.

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2 years ago

log2^2x=log8 + logx ( tak log to base 2 )

2xlog2 = 3log2 + logx

2x=3 + logx

2x = 3 + logx/log2 ( change log to base x)

2x = 3 +1/log2 ( log nw in base x)

2x log2/log2= 3 log2/log2 + 1/log2 (as log2/log2=1)

2x log2/log2=(3log2 + 1)/log2 (simplify rhs)

2x log2= 3logx +1

2xlog2 -3log2=1

log2(2x-3)=1 ( rem log is in base x)

2(2x-3)=x

4x-6=×

3x=6

x=2 (as required)

if any mistake , error or mathz error feel free to corect me. nobody is an highland of knwledge