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Do you know the solution to this equation? 109 (exponent 7)=429 (exponent 'a')

 Oct 19, 2014

Best Answer 

 #1
avatar+118612 
+10

$$109^7=429^a\\\\
log109^7=log429^a\\\\
7log109=a*log429\\\\
a=\frac{7log109}{log429}$$

 

$${\frac{{\mathtt{7}}{\mathtt{\,\times\,}}{log}_{10}\left({\mathtt{109}}\right)}{{log}_{10}\left({\mathtt{429}}\right)}} = {\mathtt{5.417\: \!746\: \!197\: \!792\: \!286}}$$

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 Oct 19, 2014
 #1
avatar+118612 
+10
Best Answer

$$109^7=429^a\\\\
log109^7=log429^a\\\\
7log109=a*log429\\\\
a=\frac{7log109}{log429}$$

 

$${\frac{{\mathtt{7}}{\mathtt{\,\times\,}}{log}_{10}\left({\mathtt{109}}\right)}{{log}_{10}\left({\mathtt{429}}\right)}} = {\mathtt{5.417\: \!746\: \!197\: \!792\: \!286}}$$

Melody Oct 19, 2014

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