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

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.

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### 4 Answers

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)

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.

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)]**

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)