# What are we learning ?

[/et_pb_text][et_pb_text _builder_version="4.1" text_font="Raleway||||||||" text_font_size="20px" text_letter_spacing="1px" text_line_height="1.5em" background_color="#f4f4f4" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px" box_shadow_style="preset2"]Competency in Focus: Powers of Numbers This problem from American Mathematics contest (AMC 8, 2013) is based on basic  algebra and Powers of Numbers.[/et_pb_text][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3"]

# Next understand the problem

[/et_pb_text][et_pb_text _builder_version="4.1" text_font="Raleway||||||||" text_font_size="20px" text_letter_spacing="1px" text_line_height="1.5em" background_color="#f4f4f4" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px" box_shadow_style="preset2"]If $3^p + 3^4 = 90$$2^r + 44 = 76$, and $5^3 + 6^s = 1421$, what is the product of $p$$r$, and $s$?[/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version="4.0"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||"][et_pb_accordion open_toggle_text_color="#0c71c3" _builder_version="4.1" toggle_font="||||||||" body_font="Raleway||||||||" text_orientation="center" custom_margin="10px||10px"][et_pb_accordion_item title="Source of the problem" open="on" _builder_version="4.1"]American Mathematical Contest 2013, AMC 8 Problem 15[/et_pb_accordion_item][et_pb_accordion_item title="Key Competency" _builder_version="4.1" open="off"]Basic algebra and Powers of Numbers[/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="4.1" open="off"]4/10[/et_pb_accordion_item][et_pb_accordion_item title="Suggested Book" _builder_version="4.0.9" open="off"]Challenges and Thrills in Pre College Mathematics Excursion Of Mathematics

[/et_pb_text][et_pb_tabs _builder_version="4.1"][et_pb_tab title="HINT 0" _builder_version="4.0.9"]Do you really need a hint? Try it first![/et_pb_tab][et_pb_tab title="HINT 1" _builder_version="4.1"]First, we're going to solve for $p$Start with $3^p+3^4=90$. Then, change $3^4$ to $81$. Subtract $81$ from both sides to get $3^p=9$ .Now we can write 9 as $3^2$ .So, from here we can say that p=2.[/et_pb_tab][et_pb_tab title="HINT 2" _builder_version="4.1"]Now, solve for $r$. Since $2^r+44=76$$2^r$ must equal $32$,  and 32 can be written as $2^5$ .So from here we have r=5.[/et_pb_tab][et_pb_tab title="HINT 3" _builder_version="4.1"]Similarly now, solve for $s$$5^3+6^s=1421$ can be simplified to $125+6^s=1421$ which simplifies further to $6^s=1296$=$6^4$ , which gives s=4.[/et_pb_tab][et_pb_tab title="HINT 4" _builder_version="4.1"]Lastly, $prs$ equals $2*5*4$ which equals $40$. So, the answer is 40.[/et_pb_tab][/et_pb_tabs][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" min_height="12px" custom_margin="50px||50px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3"]