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Question

log2+ 16log16/15+12log25/24+7log81/80=1

 

 

2) log₂x+log₄(x+2)=2

Answer 1

Hi Prasenjeet,

 

LHS=log2+ 16log16/15+12log25/24+7log81/80

     

      = log2+ log(16/15)¹⁶  +log(25/24)¹²  +log(81/80)^{7 ..........using log rule} mloga=log(a)^m

 

      = log 2* (16/15)¹⁶   *(25/24)¹²  *(81/80)⁷

 

      =log 2*( 2^4/3*5)¹⁶  *(5^2/2^3*3)¹² *(3^4/2^4*5)⁷

 

      =log (2*2^64/3^16 *5^16)* (5^24 /2^36*3^12)*(3^28/2^28*5^7)

 

      = log 2^65 *5^24*3^28/ 3^28 *2^64*5^23

 

      =log2*5 = log10=1 = RHS

 

2)   LHS=  log2x+log4(x+2)   

 

            =log x + log (x+2)/log 4..... by change of base rule

 

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

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

                

              =log x*√x+2

Assume that x=2, log 2*√2+2= log₂ 4 =2 = Rhs

 

or log₂ x*√x+2 = 2

       using rule, log_m x= y then m^y =x

       2² = x(√x+2)

       squaring on both sides,

       16 = x^2(x+2)

        x^3+ 2x^2-16=0

        x^3 -2x^2 + 4x^2 -8x+8x-16=0

       x^2(x-2) + 4x(x-2) +8( x-2) =0

        x=2

For x=2, it is proved

 

Meena from Strongsville,OH

 

                                     

Answer 2

log 2 +16 log 16/15 +12 log 24/25 + 7 log 81/80  = 1

 

 Use the the properties of log functions:

 

  Log A + log B = log AB

 

   Log Aⁿ = n Log A    log A/B = LogA - Log B

 

 log 2 + 16( 4 log 2 - log3 - log5 ) +12 ( 3log2 +log 3- 2log5) + 7 ( 4log3 -4log2- log5)=

 

    log2 + 64log2 -16log5 - 16log3 + 24 log5 - 36log2 -12 log3 + 28 log 3 - 28 log2 -7 log5 =

 

     log 5 + log 2 =

 

     log ( 5 . 2 ) = log 10 = 1

 

   log₂ X + log₄( X +2 ) =2

 

    log ₂ X + log ₂ ( X +2 ) =2

 

    log X( X +2) = log ₂4

 

     X^2 + 2X = 4

 

      X^2 +2X - 4 = 0

 

     (X + 1 ) ^2 = 5

 

      X = -1 ±√5

 

      Only √5 - 1 is acceptable

Answer 3

Hi, Prasenjeet!

 

Proving an equation in mathematics almost always means to simplify each side individually and try to get them to equal the same thing.  With that being said, let's take a look at the two questions you listed above:

 

1.) log2 + 16*log(16/15) + 12*log(25/24) + 7*log(81/80) = 1

 

The right side is already simplified as much as possible, so let's focus on simplifying the left side of the equation.  Rather than evaluating the individual logarithmic expressions on the left side, let's apply log rules to simplify the left side into one logarithmic expression. (Reminder: log(a*b)=loga + logb , log(a/b) = loga - logb , a*logb = logb^a)

 

First, let's apply the exponent rule to as many of the logarithmic expression as we can:

 

log2 + log((16/15)¹⁶) + log((25/24)¹²) + log((81/80)⁷) = 1

 

Now let's use the product rule to combine all of the logarithmic expressions into one:

 

log(2*(16/15)¹⁶*(25/24)¹²*(81/80)⁷) = 1

 

Simplify the numbers within the logarithmic expression (use your calculator - those are some big numbers!), and the expression becomes:

 

log(10) = 1

 

Evaluate log(10).

 

1=1

 

The end! :)

 

Now, let's look at your second question.  I'm assuming that this question refers to solving for x rather than solving, given that proving this equation would be kind of silly because x could be whatever you need it to be in order to make the statement true.  So let's focus on solving for x:

 

log₂x + log₄(x+2) = 2

 

The first step is going to be to try to combine the two logarithmic expressions into one logarithmic expression on the left side of the equation.  However, the two expressions do not have the same base, and so you can't combine them as is.  Let's try to rewrite them in such a way as to have the same base:

 

log₂x = y --> 2^y = x (converting into exponential form)

The base of the other log expression is 4, and we know that 2² = 4.  Let's try squaring both sides of the exponential form written above:

 

(2^y)² = x²

2^2y = x²

(2²)^y = x²

4^y = x² --> log₄(x²) = y (converting back into logarithmic form)

 

So what we just showed is that log₂(x) = log₄(x²).  Now we can use the log expression with the base of 4 instead of the log expression with the base of 2, that way both logarithmic expression in our original equation are the same base.

 

log₂x + log₄(x+2) = 2

log₄(x²) + log₄(x+2) = 2

 

Now we use the product logarithmic rule to combine both log expressions into one:

 

log₄(x²*(x+2)) = 2

log₄(x²(x+2)) = 2

 

Convert this expression into exponential form.

4² = x²(x+2)

16 = x³+2x

 

Solve for x.  I graphed the equation below on my calculator and looked at the graph.

0 = x³+2x-16

x = 2.256

 

Hope that you found this helpful!

 

Patty