Compute : 2^{log5_3} - 5^{log2_3} In latex : http://latex.codecogs.com/gif.latex?2^{log5_3}&space;-&space;5^{log2_3}

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joshmrusso | Student, Undergraduate | (Level 1) eNoter

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To answer your question let's obtain the same base for both of the exponents in the problem.

First start with 2^(log5_3)

2^(log5_3) = X  <=>  logX_2 = log5_3      {definition of log}

Then change the base of the log

(logX_3)/(log2_3) = log5_3        {change of base of log}

Then solve

logX_3 = (log5_3)*(log2_3)        {multiplication}

X = 3^[(log5_3)*(log2_3)]         {inverse of log}

Now let's change the base of the second exponent.

5^(log2_3) = Y  <=>  logY_5 = log2_3       {change of base of log}

Then change the base of the log

(logY_3)/(log5_3) = log2_3        {change of base of log}

Then solve

logY_3 = (log2_3)*(log5_3)         {multiplication}

Y = 3^[(log2_3)*(log5_3)]          {inverse of log}

All that is left is to find the difference.

2^(log5_3) - 5^(log2_3)     {original problem}

3^[(log5_3)*(log2_3)] - 3^[(log2_3)*(log5_3)]     {substitution}

0      {subtraction}

The difference is zero because 2^(log5_3) = 5^(log2_3)

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