How to add numbers with same base but unknown exponents?












0












$begingroup$


I'm taking a university math course despite knowing better. Among the straightforward problems I encountered the following:



Simplify:



$$5^x +5^{x+2}$$



The answer is supposed to be $26*5^x$



How am I supposed to arrive at this conclusion, do I interpret this as some kind of short geometric sum?






algebra-precalculus exponentiation







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    I guess you mean base instead of name.
    $endgroup$
    – lhf
    8 hours ago
















0












$begingroup$


I'm taking a university math course despite knowing better. Among the straightforward problems I encountered the following:



Simplify:



$$5^x +5^{x+2}$$



The answer is supposed to be $26*5^x$



How am I supposed to arrive at this conclusion, do I interpret this as some kind of short geometric sum?






algebra-precalculus exponentiation







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    I guess you mean base instead of name.
    $endgroup$
    – lhf
    8 hours ago














0












0








0





$begingroup$


I'm taking a university math course despite knowing better. Among the straightforward problems I encountered the following:



Simplify:



$$5^x +5^{x+2}$$



The answer is supposed to be $26*5^x$



How am I supposed to arrive at this conclusion, do I interpret this as some kind of short geometric sum?






algebra-precalculus exponentiation







share|cite|improve this question











$endgroup$




I'm taking a university math course despite knowing better. Among the straightforward problems I encountered the following:



Simplify:



$$5^x +5^{x+2}$$



The answer is supposed to be $26*5^x$



How am I supposed to arrive at this conclusion, do I interpret this as some kind of short geometric sum?






algebra-precalculus exponentiation




algebra-precalculus exponentiation






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 7 hours ago









pwerth

2,898416




2,898416










asked 8 hours ago









MagnusMagnus

186213




186213








  • 1




    $begingroup$
    I guess you mean base instead of name.
    $endgroup$
    – lhf
    8 hours ago














  • 1




    $begingroup$
    I guess you mean base instead of name.
    $endgroup$
    – lhf
    8 hours ago








1




1




$begingroup$
I guess you mean base instead of name.
$endgroup$
– lhf
8 hours ago




$begingroup$
I guess you mean base instead of name.
$endgroup$
– lhf
8 hours ago










1 Answer
1






active

oldest

votes


















7












$begingroup$

$5^{x}+5^{x+2}=5^{x}+5^{x}cdot 5^{2}=5^{x}(1+5^{2})=5^{x}cdot 26$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I didn't understand that last step, why can $5^x(1+5^2) $ be written as $5^x*26$?
    $endgroup$
    – Magnus
    8 hours ago






  • 4




    $begingroup$
    Because $5^2=25$
    $endgroup$
    – Rhys Hughes
    8 hours ago










  • $begingroup$
    ahh allright I guess I'm just tired today.
    $endgroup$
    – Magnus
    8 hours ago











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1 Answer
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1 Answer
1






active

oldest

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active

oldest

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active

oldest

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7












$begingroup$

$5^{x}+5^{x+2}=5^{x}+5^{x}cdot 5^{2}=5^{x}(1+5^{2})=5^{x}cdot 26$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I didn't understand that last step, why can $5^x(1+5^2) $ be written as $5^x*26$?
    $endgroup$
    – Magnus
    8 hours ago






  • 4




    $begingroup$
    Because $5^2=25$
    $endgroup$
    – Rhys Hughes
    8 hours ago










  • $begingroup$
    ahh allright I guess I'm just tired today.
    $endgroup$
    – Magnus
    8 hours ago
















7












$begingroup$

$5^{x}+5^{x+2}=5^{x}+5^{x}cdot 5^{2}=5^{x}(1+5^{2})=5^{x}cdot 26$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I didn't understand that last step, why can $5^x(1+5^2) $ be written as $5^x*26$?
    $endgroup$
    – Magnus
    8 hours ago






  • 4




    $begingroup$
    Because $5^2=25$
    $endgroup$
    – Rhys Hughes
    8 hours ago










  • $begingroup$
    ahh allright I guess I'm just tired today.
    $endgroup$
    – Magnus
    8 hours ago














7












7








7





$begingroup$

$5^{x}+5^{x+2}=5^{x}+5^{x}cdot 5^{2}=5^{x}(1+5^{2})=5^{x}cdot 26$






share|cite|improve this answer









$endgroup$



$5^{x}+5^{x+2}=5^{x}+5^{x}cdot 5^{2}=5^{x}(1+5^{2})=5^{x}cdot 26$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 8 hours ago









pwerthpwerth

2,898416




2,898416












  • $begingroup$
    I didn't understand that last step, why can $5^x(1+5^2) $ be written as $5^x*26$?
    $endgroup$
    – Magnus
    8 hours ago






  • 4




    $begingroup$
    Because $5^2=25$
    $endgroup$
    – Rhys Hughes
    8 hours ago










  • $begingroup$
    ahh allright I guess I'm just tired today.
    $endgroup$
    – Magnus
    8 hours ago


















  • $begingroup$
    I didn't understand that last step, why can $5^x(1+5^2) $ be written as $5^x*26$?
    $endgroup$
    – Magnus
    8 hours ago






  • 4




    $begingroup$
    Because $5^2=25$
    $endgroup$
    – Rhys Hughes
    8 hours ago










  • $begingroup$
    ahh allright I guess I'm just tired today.
    $endgroup$
    – Magnus
    8 hours ago
















$begingroup$
I didn't understand that last step, why can $5^x(1+5^2) $ be written as $5^x*26$?
$endgroup$
– Magnus
8 hours ago




$begingroup$
I didn't understand that last step, why can $5^x(1+5^2) $ be written as $5^x*26$?
$endgroup$
– Magnus
8 hours ago




4




4




$begingroup$
Because $5^2=25$
$endgroup$
– Rhys Hughes
8 hours ago




$begingroup$
Because $5^2=25$
$endgroup$
– Rhys Hughes
8 hours ago












$begingroup$
ahh allright I guess I'm just tired today.
$endgroup$
– Magnus
8 hours ago




$begingroup$
ahh allright I guess I'm just tired today.
$endgroup$
– Magnus
8 hours ago


















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