(x+3)       (7+4x)        (5+2x)        (6+3x) 2          * 5             = 3             *  4The parts between brackets are the exponents of...

(x+3)       (7+4x)        (5+2x)        (6+3x) 2          * 5             = 3             *  4

The parts between brackets are the exponents of the parts below. You can take the logaritm of the terms and work out. Thanks for helping.

Asked on by stefabts

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hala718's profile pic

hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted on

To use exponent, we use the symbol "^"

2^(x+3) *5^(7+4x) = 3^(5+2x) + 4^(6+3x)

Since the bases and powers are different, then we will use the logarithm to solve:

log [ 2^(x+3) * 5^(7+4x)] = log { 3^(5+2x) + 4^(6+3x)]

Now we know that: log a*b = log a + log b

==> log 2^(x+3)  + log 5^(7+4x) = log 3^5+2x) + log 4^(6+3x)

Also we know that: log a^b = b log a

==> (x+2)log 2 + (7+3x) log 5 = (5+2x) log 3 + (6+3x) log 4

Now let us open the brackets:

==> x log 2 + 2log 2 + 7 log 5 + 3x log 5 = 5 log 3 + 2xlog 3 + 6log 4 + 3xlog 4

Now let us group similar:

==> xlog2 + 3x log 5 -2x log 3 - 3x log 4 = 5log3 + 6log 4 - 2log2 - 7log5

Now factor out x:

==> x ( log 2 + 3log5 - 2log3) = log 3^5 + log 4^6 - log 2^2 - log 5^7)

==> x ( log (2*5^3/ 3^2) = log (3^5*4^6/2^2*5^7)

==> x  = log (3^5*4^6/2^2*5^7) / log (2*5^3/3^2)

            = log (3^5*4^5/ 5^7)/ log(2*5^3/3^2)

 

 

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neela | High School Teacher | (Level 3) Valedictorian

Posted on

2^(x+3) * 5^(7+4x) = 3^(5+2x)*4^(6+3x). To solve for x.

We  take the logarithms  of both sides.

We use the properties of logarithms: 

log(a^m) = m*loga,

logab = loga+logb and

(x+3)log2 +(7+4x) log5 =(5+2x)log3 +(6+3x) log4. Collect x's together. And numbers on other side.

x(log2+4log5-2log3 -3log4) = 5log3 +6log4 - 3log2 -7log5

x {log 2+log5^4-log3^2-log4^3) = log3^5+log4^6- lo2^3-log5^7}

 x log {2*5^4/3^2*4^3} = log{3^5*log4^6/log2^3*lo5^7}

x log {1250/576} = log(995328/625000} = 0.202086204

x*(0.336487529) = 0.202086204

x =  0.202086204/0.336487529 = 0.600575613.

 

 

 

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

To solve the problem, we'll take logarithms, both sides of the equality:

2^(x+3) * 5^(7+4x) = 3^(5+2x) * 4^(6+3x)

log [2^(x+3) * 5^(7+4x)] = log[3^(5+2x) * 4^(6+3x)]

To go further with the equation, we'll apply the product property of the logarithms:

log [2^(x+3) * 5^(7+4x)] = log 2^(x+3) + log 5^(7+4x)

and

log[3^(5+2x) * 4^(6+3x)] = log3^(5+2x) + log 4^(6+3x)

We'll substitute the logarithm of the product by the sum of logarithms:

log 2^(x+3) + log 5^(7+4x) = log3^(5+2x) + log 4^(6+3x)

Now, we'll apply the power property of logarithms:

(x+3)*log 2 + (7+4x)*log 5 = (5+2x)*log 3 + (6+3x)*log 4

We'll remove the brackets:

x*log 2 + 3*log 2 + 7*log 5 + 4x*log 5 = 5*log 3 + 2x*log 3 + 6*log 4 + 3x*log 4

We'll move all terms in x to the left side:

x*log 2 + 4x*log 5- 2x*log 3 - 3x*log 4 = 5*log 3 + 6*log 4 - 3*log 2 - 7*log 5

We'll factorize:

x*(log 2 + log 5^4 - log 3^2 - log 4^3) = log 3^5 + log 4^6 - log 2^3 - log 5^7

We'll apply the product/quotient property of logarithms:

x*log [(2*5^4)/(9*4^3)] = log [(3^5*4^6) / (2^3*5^7)]

We'll divide by log [(2*5^4)/(9*4^3)]:

x = log [(3^5*4^6) / (2^3*5^7)] / log [(2*5^4)/(9*4^3)]

thewriter's profile pic

thewriter | College Teacher | (Level 1) Valedictorian

Posted on

We have to solve the equation:

2^(x+3)*5^(7+4x)=3^(5+2x)*4^(6+3x)

taking the logarithm of terms on both the sides of the equation and using the property that log(a*b)=loga+logb and log(a^b)=b*loga we get:

(x+3)log2+(7+4x)log5=(5+2x)log3+(6+3x)log4

=>xlog2+3log2+7log5+4xlog5=5log3+2xlog3+6log4+3xlog4

taking the terms with x to one side

x{log2+4log5-2log3-3log4}=5log3+6log4-3log2-7log5

=>x= (5log3+6log4-3log2-7log5)/(log2+4log5-2log3-3log4)

=>x=log(3^5*4^6/2^3*5^7)/log(2*5^4/3^2*4^3)

=>x=log(3^5*2^12/2^3*5^7)/log(2*5^4/3^2*2^6)

=>x=log(3^5*2^9/5^7)/log(5^4/3^2*2^5)

 

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