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How would you solve this equation? It uses logarithms.

How would you solve this equation? It uses logarithms.-example-1
User J Spratt
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\bf log_{{ a}}\left( x^{{ b}} \right)\implies {{ b}}\cdot log_{{ a}}(x)\\\\ -----------------------------\\\\ 2^(x+1)=5^(2x-1)\implies log(2^(x+1))=log(5^(2x-1)) \\\\\\ (x+1)log(2)=(2x-1)log(5)\implies (x+1)\cfrac{log(2)}{log(5)}=(2x-1) \\\\\\ now\quad \cfrac{log(2)}{log(5)}\textit{ is just a constant about }0.43068

and you can simply solve that for "x"
User Karthik Ganesan
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